2.5 (x + iy)(a + ib) = xa yb + i(xb + ya) = (az by) + i(bx + ay) = (a + ib)(x + iy). The middle = uses commutativity of real numbers.

Size: px
Start display at page:

Download "2.5 (x + iy)(a + ib) = xa yb + i(xb + ya) = (az by) + i(bx + ay) = (a + ib)(x + iy). The middle = uses commutativity of real numbers."

Transcription

1 Complex Analysis Sketches of Solutions to Selected Exercises Homework 2..a ( 2 i) i( 2i) = 2 i i + i 2 2 = 2 i i 2 = 2i 2..b (2, 3)( 2, ) = (2( 2) ( 3), 2() + ( 3)( 2)) = (, 8) 2.2.a Re(iz) = Re(i(x + iy)) = Re(ix y) = y = Im(z) 2.2.b Im(iz) = Im(i(x + iy)) = Im(ix i) = x = Re(z) 2.4 ( + i) 2 2( + i) + 2 = + 2i + i 2 2 2i + 2 = 0, ( i) 2 2( i) + 2 = 2i + i i + 2 = (x + iy)(a + ib) = xa yb + i(xb + ya) = (az by) + i(bx + ay) = (a + ib)(x + iy). The middle = uses commutativity of real numbers. 2.8.a If (u, v) is an additive identity, i.e. a complex number such that (x, y)+(u, v) = (x, y), then we can subtract (x, y) from both sides to get (u, v) = (0, 0). 2.8.b We know that (, 0) is a multiplicative identity and that if (x, y) (0, 0) then there is some (x, y) so that (x, y) (x, y) = (, 0). If (u, v) is also a multiplicative identity, then (x, y)(u, v) = (x, y) for all (x, y). So we can suppose (x, y) (0, 0) and multiply by (x, y). This results in (u, v) = (, 0). 2. z 2 +z+ = (x+iy) 2 +(x+iy)+ = x 2 y 2 +2xyi+x+iy+ = x 2 y 2 +x++i(2xy+y). If this is 0 then x 2 y 2 + x + = 0 and 2xy + y = 0. Using the hint, we can divide the second equation by y to get 2x + = 0 so x = /2. Plugging in to the other equation results in y = ± 3/2 3.a +2i + 2 i 3 4i 5i 3.5 Just multiply and simplify 5..a Picture problem 5..d Picture problem = (+2i)(3+4i) (3 4i)(3+4i) i 2 i 5 = 5+0i i 5 = +2i 2i 5 = 2/5 5.5.a Circle of radius centered at i 5.5.b Closed disk of radius 3 centered at i 5.5.c Exterior of the open disk of radius 4 centered at 4i 5.8 Multiply out the left side then factor.

2 5.9 Base case: z = z = z. Induction step: Suppose z n = z n up to some n. Then z n+ = z n z = z n z = z n z = z n+. The second = is Exercise 8. The third is the induction hypothesis. 6..a z + 3i = z + 3i = z 3i 6..b iz = ix y = y + ix = y ix = i(x iy) = i z 6..c (2 + i) 2 = 3 + 4i = 3 4i 6.2.a Re( z i) = Re(x iy i) = Re(x i(y + )) = x = Re(z). So this is the line x = b 2 z + i = 2 z + i/2 = 2 z + i/2 = 2 z i/2. So the equation is the same as z i/2 = 2. So this is a circle of radius 2 centered at i/ Write out 6.0.a If z is real then z = x + i0 = x i0 = z. If z = z, then x iy = x + iy, so 2iy = 0, so y = 0, so z is real. 6.0.b If z is real then z 2 = x 2 = z 2. If z is pure imaginary then z 2 = ( iy) 2 = (iy) 2 = z 2. Conversely, if z 2 = z 2 then (x + iy) 2 = (x iy) 2 so x 2 y 2 + i(2xy) = x 2 iy 2 i2xy. So 2xy = 2xy or 4xy = 0. This implies that either x or y is 0, so z is real or pure imaginary. 9.2.a e iθ = cos θ + i sin θ = cos 2 θ + sin 2 θ = 9.2.b e iθ = cos θ + i sin θ = cos θ i sin θ = cos( θ) + i sin( θ) = e iθ Homework 2 3i 2 9..a + = 2+2, so tan θ = 3. Therefore θ has the form π/3 + kπ for some k. 3i 4 Since z is in quadrant II, Arg(z) = 2π/3 9..b ( 3 i) 6 = (4e iπ/6 ) 6 = 4 6 e πi. So Arg(z) = π. 9.0a By de Moivre s formula, cos(3θ) = Re((cos θ + i sin θ) 3 ). But (cos θ + i sin θ) 3 = (cos θ + i sin θ)(cos 2 θ sin 2 θ + i2 cos θ sin θ) = cos 3 θ cos θ sin 2 θ 2 cos θ sin 2 θ + i[ ] = cos 3 θ 3 cos θ sin 2 θ + i[ ] 2

3 ..a z = 2i = 2e iπ/2, so the square roots are 2e iπ/4 = 2( 2/2 + i 2/2) = + i and ( + i) = i..b z = 3i = 2e iπ/3, so the square roots are ± 2e iπ/6 = ±( 2( 3/2 i/2)) = ±( 3 i)/ 2.2 z = 8i = 8e iπ/2. So z /3 = 2 and arg(z /3 ) = π/6 + k2π/3. These are 2e iπ/6 = 3 i, 2e iπ/2 = 3i and 2e i7π/6 = 2( 3/2 + i/2) = 3 i i = 6e i2π/3. So z /4 = 2 and arg(z /4 ) = π/6 + k2π/4 = π/6 + kπ/2. Converting back to rectangular coordinates gives (working counterclockwise 3 i, + 3i, 3 + i, 3i.4.a ( ) = e iπ, so ( ) /3 = e iπ/3+ik2π/3. In rectangular coordinates, these are e iπ/3 = /2 + i 3/2, e iπ =, and e i5pi/3 = e iπ/3 = /2 i 3/2.4.b 8 /6 = (8e i0 ) /6 = 2e ik2π/6. These are 2(/2 ± i 3/2), 2( /2 ± i 3/2), 2, and x x is true for all real numbers, so Re(z) Re(z) and Im(z) Im(z). For the others, 0 x 2 x 2 + y 2, so, taking square roots, x x 2 + y 2, so Re(z) z, and similarly for Im 5.3 Using the preceding exercise and the triangle inequality, Re(z + z 2 ) z + z 2 z + z 2. From the alternative form of the triangle inequality, x 3 + z 4 z 3 z 4. Since z 3 z 4, neither of these values is 0 (WHY?!). So / x 3 + z 4 / z 3 z 4. No multiply the two inequalities. Show that a set is closed according to the book definition if and only if it is the complement of an open set. Let S C. Let I be the interior points of S, E the exterior points, and B the boundary points. From the book definitions, I S and E C S, so I and E are disjoint. Furthermore, B is defined to be C (E I). So E, I, and B form a partition of C (they are pairwise disjoint and their union is C). We also observe that boundary points are those points such that every ɛ-neighborhood intersects both S and C S. So, by symmetry, the boundary points of S are also the boundary points of C S. Furthermore, by the definition of E, if z E then z contains a neighborhood in C S = E. So the points of E are the interior points of C S and similarly the points of I are the exterior points of C S. Putting this all together, we see that B S if and only if C S = E if and only if C S contains none of its boundary points if and only if C S is open (by the book definition). Note that this discussion also demonstrates the claim from class that a set S is open if and only if every point z S has an ɛ-neighborhood also contained in S: The points that have this property are exactly the interior points of S, and so saying that every point of S has this property is the same as saying that S = I, which is the same as saying that S contains no boundary points, which is the book s definition of open. 3

4 2. a. Closed disk of radius centered at 2 i, not a domain; b. points outside of circle of radius 2 centered at 3/2, is a domain; c. Half play y >, is a domain; d. line y =, not a domain; e. 45 degree wedge from the origin, origin not included; not a domain; f. Half plane x 2, not a domain 2.2 e 2.3 a 2.4 a. whole plane, b. whole plane, d. arg(z) is in [ π/4, π/4] or [3π/4, 5π/4] Homework Suppose there is a boundary point z of S that is not contained in S. Since it s a boundary point, every ɛ-neighborhood of z contains a point of S. Since z is not in S, it must in fact be that every deleted ɛ-neighborhood of z contains a point of S. So z is an accumulation point of S. But this implies that z S, a contradiction. So all boundary points of S are in S and thus S is closed. 2.9 If z 0 is a point in a domain S, then since the domain is open there is an ɛ such that the disk z z 0 < ɛ is contained in S, so certainly the deleted neighborhood 0 < z z 0 < ɛ is contained in S. Any smaller deleted neighborhood around z 0 is thus also contained in S, and any larger disk intersects S in at least one point that is not z 0. Thus every deleted neighborhood around z 0 intersects S, so z 0 is an accumulation point. 4.5 There are two such domains. One of them is bounded on the left by the right branch of the hyperbola x 2 y 2 =, on the right by the right branch of the hyperbola x 2 y 2 = 2, on the bottom by the top branch of the hyperbola xy = /2, and on the top by the right branch of the hyperbola xy =. 4.8 a) quarter disk in the first quadrant given by 0 r, 0 θ π/2, b) similar but with 0 θ 3π/4, c) upper half disk Find an equation satisfied by all the complex numbers that are taken to the line x = under the map w = z 3. Without looking it up or using any algebraic geometry you should be able to give a rough sketch of this set. x 3 3xy 2 = ; picture discussed in class 8..a Let z 0 = x 0 + iy 0 and z = x + iy. Let f(z) = Re(z) = x. Let ɛ > 0. We need to show that there is a δ such that Re(z) x 0 = x x 0 < ɛ whenever 0 z z 0 δ. Recall that for any complex number a we know that Re(a) a. So in particular it is always true that x x 0 z z 0. So if we choose δ = ɛ then z z 0 < ɛ = δ implies that x x 0 < ɛ as desired. 4

5 8..c Fix any ɛ > 0. We need to show that there is a δ such that 0 < z 0 = z < δ implies z 2 /z 0 = z 2 /z < ɛ. But from the properties of conjugates and moduli, z 2 /z = z 2 / z = z 2 / z = z. So again we can take δ = ɛ, and if z < ɛ, so is z 2 /z = z. 8.5 If z = x then f(z) = (x/x) 2 =. If z = iy, then f(z) = (iy/( iy)) 2 = ( ) 2 =. But if y = z, then f(z) = ( x+ix x ix )2 = ( 2ix2 ) 2 =. So there are points arbitrarily close to 2x 2 0 that evaluate to and also points arbitrarily close to 0 that evaluate to. So the limit cannot exist. 8.6.b Suppose lim z z0 f(z) = w 0 and lim z z0 F (z) = W 0. Let ɛ > 0. Then there are a δ and such that 0 < z z 0 < δ implies f(z) w 0 < ɛ/2 and 0 < z z 0 < implies F (z) W 0 < ɛ/2. Choose δ so that 0 < δ < δ and 0 < δ <. Then if z z 0 < δ, we have f(z) + F (z) (w 0 + W 0 ) = f(z) w 0 + F (z) W 0 f(z) w 0 + F (z) W 0 < ɛ/2 + ɛ/2 = ɛ. So δ works. Since ɛ was arbitrary, this completes the argument. Homework 4 A.II. By definition, lim z z0 f(z) = w 0 means that for all ɛ > 0 there is a δ > 0 such that f(z) w 0 < ɛ whenever 0 < z z 0 < δ. But these are exactly the same formulas as for showing that lim z 0 (f(z) w 0 ) = a We look at lim z 0 4/z 2 ((/z) ) 2. Multiplying by z 2 /z 2 we get lim z 0 4 ( z) 2 = b We use that lim z (z ) 3 = c We look at (/z) (/z 2 )+ = z z2 (+z 2, which goes to 0 as z goes to By definition, S is bounded if there is an R such that z < R for all z S. So unbounded means that for all R there is a z S with z > R. But this means precisely that every neighborhood of infinity (which as the form z > R) contains a point of S. (z+ z) 20. lim 2 z 2 2z z+( z) z 0 = lim 2 z z 0 lim z z 0 2z + z = 2z 20.8.a We look at w = Re(z+ z) Re(z) = Re(z)+Re( z) Re(z) = Re( z). When z is real, this z z z z is. When z is pure imaginary this is 0. So there can t be any limit as z Note that we re only asked about the derivative at z 0 = 0. So in this case f(z 0+ z) f(z 0 ) f( z) z = z2 / z z z = = z2. When z is real, this is ( z)2 =. When z is pure imaginary, this is ( z)2 =. If x = y, this becomes ( x i x)2 = ( ) ( ) ( z) 2 ( z) 2 i 2 ( z) 2 ( x+i x) 2 +i = ( i) = ) 2 =. So there can t be a limit as z 0. ( 2i 2 5

6 24..b z z = 2iy. So v y = 0 = u x. So f is not differentiable. 24..c u x = 2 and v y = 2xy, so f can t be differentiable unless xy =. Also u y = 0 and v x = iy 2, so f can t be differentiable unless y = 0. Since y = 0 and xy = can t both happen, f is not differentiable. 24..d f = e x cos y ie x sin y. So u x = e x cos y and v y = e x cos y. These are equal only if cos y = 0. Also u y = e x sin y and v x = e x sin y, and these are equal only when sin y = 0. Since sin y and cos y cannot be 0 simultaneously, there are no points where f is differentiable b Check that the Cauchy-Riemann equations are satisfied for f = u + iv and that the partial derivatives are all continous. Then we can write f = u x + iv x and perform the same check for this function. The result is that f = u xx + iv xx = e x cos y ie x sin(y) = f a /z = x iy. So u = x and v = y. So u x 2 +y 2 x 2 +y 2 x 2 +y 2 x = y2 x 2, u (x 2 +y 2 ) 2 y = 2xy, v (x 2 +y 2 ) 2 x = 2xy, v (x 2 +y 2 ) 2 y = y2 x 2. So the CR equations are satisfied everywhere the derivatives (x 2 +y 2 ) 2 exist, but they do not exist at 0 (in fact z = 0 is not in the domain of the function). The partials are also continuous everywhere but at 0. So we have f = u x +iv x = y2 x 2 +2xyi. (x 2 +y 2 ) 2 We can recognize the top as ( z 2 ) = z 2 and the denominator as z 4 = z 2 z 2. So the quotient is /z b u = x 2 and v = y 2, so u x = 2x, u y = 0, v x = 0 and v y = 2y. So CR implies x = y, and these are continuous everywhere, so df/dz is defined along y = x where it equals u x + iv x = 2x a f = e 4θi = (cos(4θ) i sin(4θ)). So u r 4 r 4 r = 4 cos(4θ)), u r 5 θ = 4 sin(4θ)), v r 4 r = 4 (sin(4θ)) and v r 5 θ = 4 (cos(4θ)). Thus ru r 4 r = v θ and u θ = rv r, and these are all continuous (for r > 0). So the derivative exists and is e iθ (u r +iv r ) = e iθ ( 4 cos(4θ)+ r 5 i 4 iθ 4 iθ 4 (sin(4θ))) = e (cos(4θ) i sin(4θ)) = e e 4iθ = 4 e 5iθ = 4 r 5 r 5 r 5 r 5 z b u r = e θ sin(ln r)/r, u θ = e θ cos(ln r), v r = e θ cos(ln r)/r, vθ = e θ sin(ln r). So polar CR is satisfied and the partials are continuous for r > 0 and f = e θ (u r + iv r ) = e iθ ( e θ sin(ln r)/r + ie θ cos(ln r)/r) = e θ sin(ln r)+ie θ cos(ln r) = if(z)/z. re iθ Homework c u x = e y cos x, u y = e y sin x, v x = e y sin x, v y = e y cos x. So the Cauchy- Riemann equations hold and the partials are continuous everywhere in the plane. So f is defined everywhere, and f is entire a u x = y and v y =, so u x = v y is only possible when y =. Thus no point in the plane can have a neighborhood on which f is defined, so f cannot be analytic. 6

7 26.4.a The the numerator and denominator share no common factor and the denominator is 0 at 0, i, i, so f is not defined at these points. By the quotient rule, the derivative exists at all other points. So 0, i, i are singular points and the function is analytic everywhere else It s simplest to use the polar form of the Cauchy-Riemann equations. We see that u r = /r, u θ = 0, v r = 0 and v θ =. So the polar Cauchy-Riemann equations are satisfied and the partials are continuous. Thus the function is defined on the domain (notice that the domain is carefully chosen so that θ is well defined without ambiguity. Now, from our geometrical understanding of functions, z 2 takes the open first quadrant to the upper half plane and adding one then shifts everything one unit to the right. So, as the suggestion notes, Im(z 2 + ) > 0 for z in the open first quadrant, i.e. 0 < Arg(z 2 + ) < π. Since this is in the domain of g, the composition rule tells us that g(z 2 + ) is analytic. The rest follows from the CR formula for the derivative of g and from the chain rule If f is real valued and analytic on D, then f = f, so f is also analytic on D. The result now follows from Example 3 in Section The formulas u(x, y) = c and v(x, y) = c 2 determine curves that locally take the for y = y(x). As noted in the suggestion, we have from the multivariable chain rule dy (differentiating u(x, y) with respect to x) that u x + u y = 0 and v dy dx x + v y = 0. dx Find all real values of a, b, c, d so that ax 3 + bx 2 y + cxy 2 + dy 3 is harmonic. Let u = ax 3 + bx 2 y + cxy 2 + dy 3. Then u x = 3ax 2 + 2bxy + cy 2 and u xx = 6ax + 2by. Similarly, u y = bx 2 + 2cxy + 3dy 2 and u yy = 2cx + 6dy. So u is harmonic if and only if 6a = 2c and 2b = 6d, i.e. c = 3a and b = 3d. So any function of the form u = ax 3 3dx 2 y 3axy 2 + dy 3 is harmonic. Show by hand that u = x 3 3x 2 y 3xy 2 +y 3 is harmonic. Find a v so that f = u+iv is entire (hint: use the Cuachy-Riemann equations). u x = 3x 2 6xy 3y 2 and u xx = 6x 6y. u y = 3x 2 6xy + 3y 2 and u yy = 6x + 6y. So u is harmonic. By the Cauchy-Riemann equations, to find our desired v we need v x = u y = 3x 2 + 6xy 3y 2 and v y = u x = 3x 2 6xy 3y 2. Integrating v x with respect to x, we see that we must have v = x 3 + 3x 2 y 3xy 2 + g(y) for some function g depending only on y. Taking the y derivative of this, we must have v y = 3x 2 6xy + dg (y). Comparing with dy the v y we must have from the Cauchy-Riemann equations, we see that dg (y) = dy 3y2. So g(y) = y 3 + C. So we see that v = x 3 + 3x 2 y 3xy 2 y 3 + C satisfies the Cauchy-Riemann equations with u, and everything in sight is continuous. So any such v works. Homework b e 2+πi 4 = e /2 e πi/4 = e(cos π/4 + i sin π/4) = e(/ 2 + i/ 2) = e/2( + i) 7

8 30.3 e z = e x (cos y i sin y). So u x = e x cos y, u y = e x sin y, v x = e x sin y, v y = e x cos y. If CR holds, then cos y = cos y, so cos y = /2 and sin y = sin y, so sin y = /2. Since sin y and cos y are never both /2 for the same y, e z cannot be differentiable anywhere e z2 = e x2 y 2 +i2xy = e x2 y 2 e i2xy = e x2 y 2 = e x2 +y 2 e y2 = e z2 e y2. Since y 2 0 we hace e y2. So e z2 e z2 e y a e z = 2 implies that e x (cos y + i sin y) = 2. Since 2 is real, sin y = 0, so y is a multiple of π. Since 2 < 0, we must have cos y < 0, which means that is an odd multiple of π, i.e. y = (2n + )π. Then e z = e x = 2, so e x = 2, which implies that x = ln As x, e z moves toward 0 along a ray from the origin. As y, e z moves clockwise around a circle of radius e x centered at the origin an infinite number of times for sin z 2 Using (3), sin z 2 = sin 2 x cosh 2 y+cos 2 x sinh 2 y = sin 2 x cosh 2 y sin 2 x sinh 2 y+ sin 2 x sinh 2 y + cos 2 x sinh 2 y = sin 2 x(cosh 2 y sinh 2 y) + sinh 2 y(sin 2 x + cos 2 x) = sin 2 x + sinh 2 y. ( ) 2 ( ) 2 Show that sin 2 z + cos 2 z = sin 2 z + cos 2 e z = iz e iz 2i + e iz +e iz 2 = e 2iz +e 2iz e 2iz +e i2z +2 = 4/4 = 4 Show that e z = e z : e z = e x cos y + ie x sin y = e x cos y ie x sin y = e x cos( y)+ie x sin( y) = e x e iy = e x iy = e z. Show that sin z = sin z and cos z = cos z. Using the last problem and basic properties of conjugates, sin z = e i z e i z = ei z e i z = sin z. cos is similar. 2i 2i 33..a ei = e and arg( ie) = π/2 + 2πn, so Log( ei) = ln e + iθ = + i( π/2) 33.2.c + 3i = 2 and arg( + 3i) = 2π/3+2πn, so log( + 3i) = ln 2+i(2π/3+ 2πn) 33.4 With the chosen branch, log(i 2 ) = log( ) = πi, while 2 log i = 2(i5π/2) = i5π 33.9 In general, we have e z = e x+iy = e x e iy, so e z = e x and arg(e z ) = y + 2πn. If we use the branch α < θ < α + 2π for log, then we have that log(e z ) = ln e x + iθ, where Θ is the value of arg(e z ) = y + 2πn with α < Θ < α + 2π. But with the assumption, this is precisely y. So log e z = x + iy = z a By definition/branch cuts, Log(z) is analytic so long as z does not lie on the part of the real axis with x 0. Thus, since the composition of anlaytic functions is analytic where it is defined (and Log z and z i are analytic) Log(z i) is thus analytic so long as z i is not on the non-positive x-axis, which is equivalent to z not being on the line y = with x 0. 8

9 33. One way to do this is by direct computation of derivatives. Alternatively, ln(x 2 + y 2 ) is the real part of 2 log z for any branch cut. For any branch cut, 2 log z is analytic in its domain, and so its real part is harmonic in that domain, which covers all of the plane except one ray. If we choose a different branch cut, then we see that ln(x 2 + y 2 ) is also harmonic on the ray, except for at 0. Since being harmonic is a local property and we have verified it at all non-zero points of the plane, the function is harmonic on C {0}. Homework Let z = 2/2 + i 2/2, and let z 2 = i. Then z /z 2 = 2/2 i 2/2, so Log(z / z ) = i3π/4. But Log(z ) Log(z 2 ) = i3π/4 ( iπ/2) = i5π/ a ( + i) i = e i log(+i) = e i(ln 2+i(π/4+2πn)) = e π/4+2πn e i ln 2 = e π/4+2πn i(ln 2)/2 e 36.2.a Principal value of ( i) i = e i Log( i) = e i( πi/2) = e π/ c Principal value of ( i) 4i = e 4i Log( i) = e 4i(ln 2 πi/4) = e π+i2 ln(2) = e π (cos(2 ln 2) + i sin(2 ln 2)) 36.6 If a is real then z a = e a log z = e a(ln z )+ai arg(z) = e a ln z e ai arg z = e a ln z = e ln z a = z a. But if we take the principal value of z a then since z is a positive real number we have z a = e a Log z = e a ln z as real numbers, and this is the usual real number z a, which is also positive. So z a = z a a Using principal values, we have z c z c 2 = e c Log z e c 2 Log z. As noted in Section 35, when we use principal values e z is exactly the function we re used to working with, so, in particular, e c Log z e c 2 Log z = e c Log z+c 2 Log z = e (c +c 2 ) Log z, which is the principal branch of z c +c a 0 ( + it)2 dt = 0 + 2it t2 dt = t + it 2 t 3 /3 0 = + i /3 (0) = i 42.2.c π/6 e i2t dt = ei2t 0 2i π/6 0 = eiπ/3 2i = ( + i 3 = 3+i 2i π 0 e(+i)x dx = e(+i)x π +i 0 = e(+i)π = eπ = +i +i 2 ( eπ )( i). So π 0 ex cos x dx = 2 ( eπ ) and π 0 ex sin x dx = 2 (eπ + ). 43..a Let τ = t. Then dτ = dt and a b w( t)dt = a b w(τ)dτ = b a w(τ)dτ 46..a z = 2ie iθ so C 2πi 2 = 4 + 2πi z+2 z dz = π 0 2e iθ +22ie iθ dθ = π 2e iθ 0 2ieiθ + 2i dθ = 2e iθ + 2iθ π 0 = b Take z(t) = t, 0 t 2. Then C z dz = 2 0 t dt = t2 /2 t 2 0 = = Our parametrization is z = t + it 3 but we need to consider two pieces, t 0 and 0 t. We also have z = + 3t 2 i. The first part gives us 0 ( + 3t2 i) dt = t+t 3 i 0 = ( i) = +i. The second part is 0 4t3 (+3t 2 i) dt = 0 4t3 +2t 5 i dt = t 4 + 2t 6 i 0 = + 2i. So altogether we have + i + + 2i = 2 + 3i 9

10 46.5 Given any z(t), a t b, we have dz = b dz dt = C a dt z b a = z(b) z(a) = z 2 z 46.6 Note that the contour only has an endpoint on the branch cut. So we use z = ie iθ and the integral is π 0 ei Log z ie iθ dθ = π 0 ei(iθ) ie iθ dθ = π 0 e θ ie iθ dθ = π 0 eiθ θ idθ = π 0 e(i )θ idθ = ie(i )θ π i 0 = ieπi π i = ie π i = ( i)( ie π i) = ( +i)(e π +) i i 2 2 Homework Let w(t) = f(z(t)) = u(x(t), y(t))+iv(x(t), y(t)). Then by the chain rule, dw = u dx + dt x dt u dy + i v dx + i v dy. Writing u y dt x dt y dt x, u y, v x, v y, x, y for the appropriate derivatives and using Cauchy-Riemann, this is u x x +u y y +i(v x x +v y y ) = u x x v x y +i(v x x +u x y ) = (u x + iv x )(x + iy ) = df dz dz dt as desired. 47..a On part of the circle of radius 2, z+4 z +4 = 2+4 = 6 and z 3 z 3 = 7. The length of the quarter circle is 2π 2/4 = π. So by the Theorem in this section, the integral is 6π/ As noted in the suggestion, the minimum z on the curve occurs at the midpoint /2 + i/2 and so is /2. Thus /z 4 /( /2) 4 = 4. The length of the line segment is 2 by the Pythagorean theorem. So the integral is On the circle of radius R we have that 2z 2 2 z 2 + = 2R 2 + while z 4 +5z 2 +4 = z z 2 + z 2 4 z 2 = (R 2 4)(R 2 2z ). So 2 2R 2 + z 4 +5z 2 +4 (R 2 4)(R 2 ) 2R and the integral is 2 + πr. As R this expression goes to 0. (R 2 4)(R 2 ) 47.7 x + i x 2 cos θ = x 2 + ( x 2 ) cos 2 θ. Since x 2 0 by the assumption x, we have ( x 2 ) cos 2 θ x 2. So x + i x 2 cos θ x 2 + x 2 =. So (x + i x 2 cos θ) n = x + i x 2 cos θ n n =. So the modulus of the integral is π and therefore P n (x). 49. For n a nonnegative integer, z n is entire with antiderivative F (z) = zn+ n+ plane. So a contour integral from z to z 2 of z n is F (z 2 ) F (z ) = zn+ 2 z n+. n+ in the whole 49.2.b 2 sin(z/2) is an antiderivative of cos(z/2). So the integral is 2(sin(π/2+i) sin(0)) = 2 sin(π/2 + i) = ei(π/2+i) e i(π/2+i) = e +iπ/2 e iπ/2 = e i e( i) = e + e i i i 49.3 Since the case n = 0 is omited, (z z 0 ) n has antiderivative (z z 0) n on the domain n C {z 0 } (if n > 0 this is an antiderivative on all of C). As long as a contour does not pass through z 0, the integrand is defined and continuous on the contour. So why the Theorem in Section 48, the integral is 0 when the contour is also closed As observed in the text, the branch of z /2 using π/2 < θ < 5π/2 is defined at all points of any contour that lies below the real axis except for its endpoints at 3 and 3, and for all points on such a contour the values of this branch aree with those of the branch 0 < θ < 2π, except at the point 3. But as one point does not affect the value 0

11 of the integral, we can use the π/2 < θ < 5π/2 branch to compute. Furthermore, this branch of z /2 has antiderivative z 3/2 /(3/2) (with the same branch choice) on a domain containing the contour. So by the Theorem in Section 48, the contour integral is 2 3 (33/2 ( 3) 3/2. For this branch, 3 3/2 = (3e 2πi ) 3/2 = 3 3/2 e 3πi = 3 3/2 while ( 3) 3/2 = (3e πi ) 3/2 = 3 3/2 e i3π/2 = i3 3/2. So the integral is 2/3 ( 3 3/2 + i3 3/2 ) = 2( 3 + i 3) As in the suggestion, if we consider the branch of z i with π/2 arg z < 3π/2 to write z i = e i log z then this agrees with e i Log z on the entirety of any contour that lies about the real axis (except for its endpoints). So the integral is the same treating z i either way. But now using our branch, z i has antiderivative z i+ /(i + ) (using the same branch to define z i+ ) and so by the Theorem in Section 48, the integral is ( i+ ( ) i+ )/(i+). Using our branch, we compute i+ = e (i+) log = e (i+)(0+0i) = e 0 =, while ( ) i+ = e (i+) log( ) = e (i+)(ln +iπ) = e π+iπ = e π. So the integral is +e π = +e π ( i) i+ 2 Homework a This function is analytic except at z = 3. So it is analytic on and inside the circle, so the integral is c This function only fails to be analytic at 2± = ± i. These points are outside the disk, so by Cauchy-Goursat the integral is f f fails to be analytic at the points {x + iy x 2, y = 0}. These don t intersection the unit disk, so by Cauchy-Goursat the integral is a This function is analytic except where z = ± 3i. So, for example, f is analytic on the domain z > 0.6. Thus the Theorem (more precisely the Corollary) from Section 53 applies z 0 = 2 + i is in the interior of the rectangle described, and for all n the function (z 2 i) n is analytic on C {z 0 }. So by the Corollary of Section 53, the integral on C 0 equals the integral on C. The result follows from the C 0 integral computations given We can t use Cauchy-Goursat here because the square root isn t well defined at 0, so the hypotheses of the theorem aren t satisfied. However, we compute the pieces. On the semicircle we use z = e iθ so z = ie iθ and we have π 0 eiθ/2 ie iθ dθ = i π 0 ei3θ/2 dθ = e i3θ/2 2/3 = (2/3)(e i3π/2 ) = (2/3)( i ) = 2/3 i2/3. For the part on the positive x axis we have 0 t dt = (2/3)t 3/2 0 = 2/3. For the negative real part we have 0 te iπ/2 dt = i2/3. Adding these three pieces, we get This comes from Green s Theorem. With f = x iy, Green s Theorem (see (4) in Section 50) says that u ivdz = v C R x u y +i(u x +v y )da = 0 0+i(+) da = R R2i da = 2i Area(R). Homework 0

12 57..a 2πie πi/2 = 2πi( i) = 2π 57..b 2πi cos(0) = πi/4 57..c z = z/2 2z+ z+/2 so the answer is 2πi( /4) = πi/2 57..e 2πi d dz tan(z/2) = 2πi sec2 (z/2)/2. So the answer is πi sec 2 (x 0 /2) 57.2.a z = (z + 2i)(z 2i). 2i is in the circle but 2i isn t. So the integral is 2πi evaluated at 2i. So the answer is 2πi/4i = π/2 times z+2i 57.3 g(2) = 2s 2 s 2 ds, so g(2) = 2πi(2(2) 2 2 2) = 2πi4 = 8πi. If z > 3 then C s 2 the integrand is analytic on and inside the contour so g(z) = If z 0 is outside the contour, both sides are 0. If z 0 is inside the contour and C is positively oriented then from the Cauchy formulas both sides are f (z 0 ), using that f analytic implies that f is analytic. If C is negatively oriented, then the integrals reverse signs but are still equal From the integral formula, the integral is 2πie a0 = 2πi. In terms of θ, we let z = e iθ so that z = ie iθ. Then the integral is π π ea(cos θ+i sin θ) ie iθ /e iθ dθ = i π π ea cos θ e ia sin θ dθ = i π π ea cos θ (cos(a sin θ)+i sin(a sin θ))dθ. Since e a cos θ sin(a sin θ) is an odd function, its integral from π to π is 0. So we get i π π ea cos θ cos(a sin θ)dθ = 2πi. Also e a cos θ cos(a sin θ) is even so its integral from π to 0 is equal to the integral from 0 to π. So π 0 ea cos θ cos(a sin θ)dθ = π 57.0 Let z 0 be a point in the plane, and let C R be the circle of radius R around z 0. The points on C R have the form z 0 + Re iθ. Then on C R we have f(z) A z 0 + Re iθ A( z 0 + R). So f (z 0 ) 2A( z 0 + R)/R 2. As R goes to, we see that f (z 0 ) = 0. So f (z) is constant, i.e. f (z) = a. Since f is an antiderivative of f, this implies f(z) = a z + C. But f(0) must be 0 for f(z) A z to hold, f(z) = a z. 59. If f is entire, so is e f = e u+iv = e u e iv. So if u is bounded, then so is e f = e u. So then e f is constant. So e f = e u is constant. So u = lne u is constant Let R be the region z. Then if f(z) = z, f(z) = z = 0 at z = 0, but f(z) = z > 0 for z 0. So f has a minimum in the interior of R a Just multiply and cancel 59.8.b By part a, we can write z k z k 0 = (z z 0 )P k (z), where P k is a polynomial of degree k. P (z) P (z 0 ) = a 0 + a z + a 2 z a n z n a 0 a z 0 a 2 z 2 0 a n z n 0 = a (z z 0 ) + a 2 (z z 0 ) a n (z n z n 0 ) = a (z z 0 ) + a 2 (z z 0 )P (z) + + a n (z z 0 )P n (z) = (z z 0 )(a + a 2 P (z) + + a n P n (z) = (z z 0 )Q(z) When P (z 0 ) = 0 we see that P (z) = (z z 0 )Q(z), as required. 2

13 Suppose f(z) is entire and f(z) for all z. Show that f is constant. Since f(z), we have, so is bounded. It is entire since f(z) is never f(z) f(z) 0. So /f(z) is constant by Liouville s Theorem. So f(z) is also constant. Let R be a closed bounded region of the plane. Suppose f and g are continuous on R and analytic in the interior of R. Show that if f = g on the boundary of R then f = g on all of R. Consider f g, which is 0 on the boundary. By the corollary to the maximum modulus principle, the maximum of f g is on the boundary of R, so f g must be 0 on all of R. So f = g. Technically that corollary requires f g not be constant, but if f g is constant, then since it s 0 on the boundary it s 0 everywhere in R so again f = g. What is the maximum of e iz2 on the disk z. From the Maximum Modulus Principle, we know that the maximum must be on the boundary. So we consider e iz2 = e i(x2 y 2 +2ixy) = e 2xy on the circle. This will have its maximum where 2xy has its maximum on the circle. Using Calc III methods (either parametrize the curve or use Lagrange multipliers), the maximum will be where y = x on the circle, i.e. (/ 2)( i) and (/ 2)( + i). Then e 2xy = e. 6. We want ( n 2 + i) i = n 2 < ɛ. This will be true so long as n > / ɛ Show directly from the definitions that if n= a n = A and Homework then n= (a n + b n ) = A + B. Let S m = m n= a n and T m = m n= b n = B n= b n. Then by definition for all ɛ > 0 there are M and M 2 so such that S m A < ɛ for m > M and T m B < ɛ for m > M 2. By taking the larger of M, M 2, we see there is a single M so that S m A < ɛ and T m B < ɛ simultaneously for m > M. Now let U m = S m +T m = m n= (a n +b n ) (we can do this because these sums are finite). Then for M > m, U m (A + B) = S m + T m (A + B) = S m A + T m B S m A + T m B. If we choose M so that S m A, T m B < ɛ/2 for m > M, then we have U m (A + B) < ɛ for m > M. This shows that U m converges to A + B as desired b e z = ee z = e (z ) n. This applies for all z C. n! 65.4 cos z = sin(z π/2) = (z π/2) 2n+. This holds everywhere in C. (2n+)! 65.8.a cos z = eiz +e iz = ( (iz) n + ( iz) n ) = ( (iz) n + ( ) n (iz) n ). Using that the 2 2 n! n! 2 n! n! terms are negatives of each other for n odd and equal to each other for n even this becomes ( 2(iz) 2n ) = (iz) 2n 2 (2n)! (2n)! 65.9 f(z) = ( )n (z 2 ) 2n+ (2n+)! = (i) 2n z 2n (2n)! = ( ) n z 2n (2n)!. = z4n+2 ( )n. Since this only has powers of the form (2n+)! z 4n+2, all of the terms z 4n, z 4n+, z 4n+3 must have trivial coefficients, so f (4n) (0) = 0 and similarly for the others. Notice that 4n+ and 4n+3 together give all the positive odd integers, so we can restate that condition as vanishing for all 2n +. 3

14 z Find a Maclaurin series for 3. On what set does this converge? z 3 = z 2 +6 z 2 +6 z 3 = z3 6 6 ( z2 6 )n = z2n+3 ( )n. This converges when z 2 /6 <, i.e. (6) n+ z2 6 when z < 4 Find a Taylor series for z z centered at z = 3. What is the region of convergence? z z = z (z 3) z 3 = 2 (z 3) (z 3) = z 3 2 z z = z 3 ( ) n z ( ) n z ( ) n+ z 3 3 (z 3) n = + 2 ( 2) n+ ( ) n z 3 = (z 3) n 2 ( 2) n n= n= = ( + 3 ) (z 3) n 2 ( 2) n n= = n= n= (z 3) n 2 ( 2) n = (z 3) n ( 2) n+ This converges for z 3 < sin z = z z3 + z5 z7 + so sin z 2 = z 2 z6 + z0 z4 + and 3! 5! 7! 3! 5! 7! z z2 + z6 z0 + 3! 5! 7! sin z2 z 4 = 4

15 65. 4z z = 2 4z z 4 = ( z n 4z 4) z n = 4 n+ = 4z + z n 4 n+2 Find Laurent expansions about z 0 = 0 for z > 2. When 0 < z < 2 we have When z > 2, z 3 4z = z 3 4 z 2 = z (just for D 2 ) f(z) = + z n+ z n = 2 n+ z 3 4z = = z 3 4z 4z ( z 2 )2 4z ( 4 z 2 ) n = = = (z )(z 2) z z 2 z z z n + 2 n+ n= z n on the regions 0 < z < 2 and ( z2 4 )n = z 2n. 4 n+ 2 4 n z 2n+3 z 2 = z + z n z n 2 = 2 n Let C be the contour z = 2 oriented positively. Compute z cos(/z) dz. We have C the Laurent series z cos(/z) = z ( ) ( n ) 2n (2n)! z = z cos(/z) = ( ) n. So (2n)! z 2n b = /2. Thus the integral is 2πi( /2) = πi. Homework Differentiating both sides of the first equation, we get ( z) 2 )z n 2. Differentiating again, ( z) 3 is all valid only within the circle of convergence z < = n= n(n + )zn = 72.4 Using the Taylor series for cos z, which is entire, cos z = n= = n= nzn = (n+ (n + )(n + 2)zn. This ( ) z 2n (2n)! z 2n. So ( cos (2n)! z)/z2 = z 2n 2 n=. At z = 0, this is /2, so this series (2n)! represents f(z) on all of C. Since the series for cos z converges for all z, our new series also converges for all z (for each fixed z, multiplying by is multiplication by a z 2 constant and so doesn t affect the convergence). Since the series converges everywhere, it represents an entire function by the corollary in section Let C be a contour from to z where z satisfies z <. Then we can integrate both sides along this contour. On the left we get dw = Log C w w z = Log z Log = Log z. on the right we have C ( )n (w ) n dw = C ( )n (w ) n dw = ( )n (w ) n+ z n+ = ( )n (z ) n+ = n+ n= ( )n+ (z ) n. n 72.7 Away from z = on the described domain, f(z) is analytic by the analyticity of Log and the quotient rule. Inside the circle z < (and only here), f(z) = 5 =

16 n= ( )n+ (z ) n = n ( )n (z ) n. Note that this agrees with f() =, and n+ the series absolutely converges inside the circle by comparison to the geometric series z n. So f(z) is analytic at also. Show that if f is analytic in a domain except for finitely many singular points then they are all isolated singularities. Let z,..., z n be the singular points. Without loss of generality, consider z. For each i >, let R i = z z i. Since there are finitely many singular points, there is a minimum R = min{r 2,..., R n }. Now choose ɛ so that 0 < ɛ < R and so that z has an ɛ neighborhood in the domain (this is possible because domains are open). Then f is analytic on the deleted ɛ neighborhood around z, so z is an isolated singular point. The argument is the same for the other singular points. 77..a z+z 2 = z = ( z + ) = + for z <. So the residue at 0 is. +z z z 77..b z cos(/z) = z( + ) = z + on the whole plane. So the residue is 2z 2 2z / c z sin z z = z (z (z z3 /6 + )) = z (z3 /6 + ) = z 2 /6 +. So the residue is a e z = ( z + z 2 /2 + ) = /z2 +. So the integral is 2πi( ) = 2πi. z 2 z 2 z 77.2.d There are singularities at 0 and 2. By partial fractions, z+ z 2 2z = 2z + 3 2(z 2). At z = 0, the second function is analytic, so the residue is /2. At z = 2, the first function is analytic so the residue is 3/2. So the integral is 2πi( /2 + 3/2) = 2πi Let C be the positively oriented circle z = 5. Compute sin z. To find the residue C (z π) 2 at π, we need to expand sin z in powers of z π. We use sin(z) = sin(z π + π) = sin(z π) cos π + cos(z π) sin π = sin(z π) = (z π) + (z π) 3 /6. So = /(z π) +. So the residue is and the integral is 2πi. sin z (z π) Looking at f(/z) we have z 2 z 2 = 4, so the integral is 8πi. 4 5(0) ( 0) 4/z 5 (/z)(/z ) = z 2 z(4 5z) z = 4 5z z( z). The residue at 0 is 77.4.a f(/z) = /z 5 = = z 2 z 2 /z 3 z 7 /z 3 z 4 the integral is 2πi. z 3 = z 4 z3n. So the residue is and 77.4.b f(/z) = =. This is analytic at 0 so the residue at 0 is 0. So the z 2 z 2 +/z 2 z 2 + integral is Consider P (/z)/q(/z). The biggest power of /z in the expression is /z m. Multiplying top and bottom by z m, we get a polynomial with non-zero constant term on the bottom and a polynomial with 0 constant term and linear term on the top. Dividing by z 2, the numerator is still a polynomial and the denominator is a polynomial with non-zero constant term. So P (/z) is analytic at 0, so its residue at 0 is 0 and the z 2 Q(/z) integral is 0. 6

17 79..a The principal part is n=2. This singularity is essential. z n n! 79..b Note that z 2 = (z + ) 2 2z = (z + ) 2 2(z + ) +. So z2 z+ So the principal part is and this is a simple pole. z+ 79..c sin z/z = z2n ( )n. So this is a removable singularity. (2n+)! = (z + ) 2 + z b e2z z 4 = z 4 ( 2n z n /n!) = z 4 ( n= 2n z n /n!). So the principal part is 2/z 3 2/z 2 8/6z. So the pole has order 3 and residue 4/3. Homework 3 Section 83. From the Maclaurin series we know that sin z has a zero of order. So by the Theorem in section 83, the residue is cos 0 = We have q(z) = cos z, q (z) = sin z, and q (z) = cos z. So q(0) = q (0) = 0 but q (0) =. So 0 is a zero of order a We can write z sec z as z/ cos z. Since f(z) = cos z is 0 as π/2+nπ but f (z) = sin z is not zero at this points, we see that cos z has zeros of degree at these points. So z from theorem, the residues are evaluated at these points. For z sin z n = π/2 + πn we have sin(z n ) = ( ) n, so the residues are ( ) n z n as desired a tan z = sin z/ cos z has singularities at ±π/2. From the theorem in section 83, the residue at these points is sin z/( sin z) = evaluated at these points, so each is just. So by the residue theorem, the integral is 2πi( ) = 4π Homework 4 z+ z+ 8..a f(z) =. So at 3i we have a simple pole with φ(z) = (z 3i)(z+3i) z+3i φ(3i) = +3i. And 3i we have a simple pole with φ(z) = z+ 6i z 3i 3i 6i. and residue and residue φ( 3i) = 8..c f(z) = z (z+/2) 3. So we have a pole of order 3 at /2 with φ(z) = z 3 /8. So the residue is φ ( /2)/2 = 3/ a ( ) /4 is not 0, so we have a simple pole at with residue ( ) /4 = (e πi ) /4 = e πi/4 = 2/2 + i b f(z) = Log z, so we have a pole of order 2 at i with φ(z) = Log z. The residue (z+i) 2 (z i) 2 (z+i) 2 is φ (i). φ = (z+i)2 (/z) (Log z)(2z+2i) (z+i) 4 so φ (i) = (2i)2 (/i) (Log i)(4i) (2i) 4 = 4i+2π = π+2i b Note that φ is not because this is undefined at 0. Rather we observe that e z ez = n= zn /n! = z zn /(n + )!. So f(z) = where g(z) = z 2 g(z) zn /(n + )! or φ(z)/z 2 with φ = /g. Thus we have a pole of order 2 and the residue is φ (0). Then φ (0) = g (0)/g 2 (0). From the series for g we have g(0) = while, differentiating term by term, g = n= nzn /(n + )!. So g (0) = /2. Thus φ (0) = /2. 7

18 8.5.a The contour only goes around the pole 4, where the residue is /4 3 = /64. So the integral is 2πi/64 = πi/ b The contour goes around both poles. The integral is 2πi times the sum of the residues. At 4 we have a simple pole with residue /( 4) 3 = /64. At 0 we have a pole of order 3 and so need φ (0)/2. Here φ = /(z + 4), so φ = /(z + 4) 2 and φ /2 = /(z + 4) 3. At 0 this is /64. So the integral is 2πi(/64 /64) = By contradiction, assume there are an infinite number of zeros. Since the contour together with the region inside it constitute a closed bounded region, by the Bolzano- Weierstrass Theorem, the set of zeros must have an accumulation point, i.e. a point so that there is a zero in every deleted neighborhood. φ(z) (z z 0 ) m This accumulation point cannot be a pole: In a neighborhood of a pole f(z) = for some φ that is analytic at z 0 and with φ(z 0 ) 0. But then if z i is a sequence of zeros of f for which z 0 is a limit point we must have φ(z i ) = 0 in this neighborhood. So by continuity we also have φ(z 0 ) = 0, a contradiction. So the accumulation point is not a pole but rather a point z 0 at which f is analytic and again by continuity f(z 0 ) = 0. But now by (the contrapositive of) the theorem in section 82, this situation can only happen is f is identically zero in a neighborhood of the accumulation point. But then the accumulation point is a zero of infinite order (and/or on the boundary), contradicting the assumption that all the zeros have finite order and are interior to C. 86. Since the integrand is even, we start with the PV integral f(x) dx. Letting f(z) = /(z 2 + ) =, this has poles at ±i. The residue at i is /2i = i/2. So (z i)(z+i) integrating around a semi-circular contour, the PV integral will be 2πi( i/2) = π and the desired integral will be π/2 if we can show that the integral around the top part of the contour goes to 0. Since z 2 + z 2, on the semi circle of radius R we ll have f(z) R 2 as R goes to infinity.. So the integral of the top semicircle is R 2 πr. This goes to The procedure is essentially the same as in the preceding problem but now f(z) = (z i) 2 (z+i) 2. We have a pole of order 2 at i, and letting φ(z) = /(z+i) 2, the residue will be φ (i). Since φ = 2/(z+i) 3, we have φ (i) = 2/(2i) 3 = /4i 3 = / 4i = i/4. So the PV integral will be 2πi( i/4) = π/2 and the desired integral will be π/4. For the top semicircle, we use that the maximum will be /(R 2 ) 2 using the triangle inequality. z 86.5 Let f(z) = 2. The poles in the upper half plane will be the simple poles at i (z 2 +)(z 2 +4) and 2i. The easiest way to find the residues here will be to use Theorem 2 in Section 83. In this case, p/q z = 2 = z2 = 2z(z 2 +4)+(z 2 +)2z 4z 3 +z fracz4z So the desired integral will be πi( i + 4i 2 +0 circle integral is R 2 (R 2 )(R 2 4) 2i ) = πi( i + 2i 4(2i) πr, which goes to 0 8 ) = πi( i 6 ) = π/6. The upper half

19 = 3(e πi/3 ) 2 3e 2πi/ First note that /(z 3 + ) has simple poles at e πi/3+2πni/3. The only one of these in the contour is e πi/3. Using the theorem from section 83, the residue there is = e 2πi/3 /3. So the contour integral gives 2πie 2πi/3 /3. The integral along the circular arc of the contour is R 3 2πR 3 which goes to 0 as R goes to infinity. For the diagonal piece, letting the contour grow to infinity, we note that we can parameterize it in the wrong direction by z(t) = te i2π/3, 0 t <. Writing out the contour integral along this piece in the wrong direction using the parametrization we dt. Not that the integral t 3 + here is the same one we re interested in evaluation and that we get along the positive x axis. However, the contribution to the contour is negative since we parameterized in the wrong direction. So altogether the contour integral has three parts: the circle part get 0 (te i2π/3 ) 3 + ei2π/3 dt = 0 that goes to 0, the part 0 t 3 + ei2π/3 dt = e i2π/3 0 dx and the diagonal part x 3 + ei2π/3 0 dx. x 3 + Adding these up and using our residue computation from earlier, we have that ( e i2π/3 ) = 0 x 3 + 2πie 2πi/3 /3. So the integral is 2πie 2πi/3 2π. This simplifies to 3( e i2π/3 ) 3 3 Homework 5 by basic arithmetic/trigonometry. Let C be the curve z =2 oriented positively. For each of the following functions f(z), determine how many times (with sign) the image f(c) winds around the origin: ) f(z) = z 3, winding number 3; 2) f(z) = z 4 /(z ) 2, winding number 2; 3) f(z) = /(z 2 + ) 2, winding number Suppose that f has a zero of degree m k at z k. Then near z k, f(z) = (z z k ) m k g(z), where g(z) is analytic at z k and f(z k ) 0. Then f = m k (z z k ) mk g(z) + (z z k ) m k g (z). Then zf /f = z(m k(z z k ) m k g(z)+(z z k ) m k g (z)) (z z k = zm ) m k k g(z) z z k + zg (z). The second g(z) summand is analytic at z k and, unless z k = 0, the first has a simple pole at z k with residue m k z k (remember that if φ(z) is analytic at z 0 then the residue of φ(z)/(z z 0 ) at z 0 is φ(z 0 )). If z k = 0 then there is no pole, but this is consistent with the contribution to the formula being m k z k = 0. Since f is analytic, the only singularities of zf /f are where f(z) = 0. So from the residue theorem, the integral is 2πi m k z k c On z =, 4z 3 = 4, while z 7 + z z 7 + z + = 3. So the full polynomial has the same number of zeros in the circle as 4z 3, which has a On z = 2, we have 9z 2 = 36 while z 4 2z 3 + z = 35. So the full polynomial has as many zeros in the circle as 9z 2, which has On the circle, cz n = c while e z = e x e = e. So since e < c by assumption, cz n and cz n e z have the same number of zeros in the circle, which is n. 9

MATH 452. SAMPLE 3 SOLUTIONS May 3, (10 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic.

MATH 452. SAMPLE 3 SOLUTIONS May 3, (10 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. MATH 45 SAMPLE 3 SOLUTIONS May 3, 06. (0 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. Because f is holomorphic, u and v satisfy the Cauchy-Riemann equations:

More information

Topic 4 Notes Jeremy Orloff

Topic 4 Notes Jeremy Orloff Topic 4 Notes Jeremy Orloff 4 auchy s integral formula 4. Introduction auchy s theorem is a big theorem which we will use almost daily from here on out. Right away it will reveal a number of interesting

More information

Math 185 Fall 2015, Sample Final Exam Solutions

Math 185 Fall 2015, Sample Final Exam Solutions Math 185 Fall 2015, Sample Final Exam Solutions Nikhil Srivastava December 12, 2015 1. True or false: (a) If f is analytic in the annulus A = {z : 1 < z < 2} then there exist functions g and h such that

More information

Complex Homework Summer 2014

Complex Homework Summer 2014 omplex Homework Summer 24 Based on Brown hurchill 7th Edition June 2, 24 ontents hw, omplex Arithmetic, onjugates, Polar Form 2 2 hw2 nth roots, Domains, Functions 2 3 hw3 Images, Transformations 3 4 hw4

More information

18.04 Practice problems exam 2, Spring 2018 Solutions

18.04 Practice problems exam 2, Spring 2018 Solutions 8.04 Practice problems exam, Spring 08 Solutions Problem. Harmonic functions (a) Show u(x, y) = x 3 3xy + 3x 3y is harmonic and find a harmonic conjugate. It s easy to compute: u x = 3x 3y + 6x, u xx =

More information

Solutions to practice problems for the final

Solutions to practice problems for the final Solutions to practice problems for the final Holomorphicity, Cauchy-Riemann equations, and Cauchy-Goursat theorem 1. (a) Show that there is a holomorphic function on Ω = {z z > 2} whose derivative is z

More information

Solution for Final Review Problems 1

Solution for Final Review Problems 1 Solution for Final Review Problems Final time and location: Dec. Gymnasium, Rows 23, 25 5, 2, Wednesday, 9-2am, Main ) Let fz) be the principal branch of z i. a) Find f + i). b) Show that fz )fz 2 ) λfz

More information

Lecture Notes Complex Analysis. Complex Variables and Applications 7th Edition Brown and Churchhill

Lecture Notes Complex Analysis. Complex Variables and Applications 7th Edition Brown and Churchhill Lecture Notes omplex Analysis based on omplex Variables and Applications 7th Edition Brown and hurchhill Yvette Fajardo-Lim, Ph.D. Department of Mathematics De La Salle University - Manila 2 ontents THE

More information

MATH 106 HOMEWORK 4 SOLUTIONS. sin(2z) = 2 sin z cos z. (e zi + e zi ) 2. = 2 (ezi e zi )

MATH 106 HOMEWORK 4 SOLUTIONS. sin(2z) = 2 sin z cos z. (e zi + e zi ) 2. = 2 (ezi e zi ) MATH 16 HOMEWORK 4 SOLUTIONS 1 Show directly from the definition that sin(z) = ezi e zi i sin(z) = sin z cos z = (ezi e zi ) i (e zi + e zi ) = sin z cos z Write the following complex numbers in standard

More information

Complex Variables. Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit.

Complex Variables. Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit. Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit. 1. The TI-89 calculator says, reasonably enough, that x 1) 1/3 1 ) 3 = 8. lim

More information

(a) To show f(z) is analytic, explicitly evaluate partials,, etc. and show. = 0. To find v, integrate u = v to get v = dy u =

(a) To show f(z) is analytic, explicitly evaluate partials,, etc. and show. = 0. To find v, integrate u = v to get v = dy u = Homework -5 Solutions Problems (a) z = + 0i, (b) z = 7 + 24i 2 f(z) = u(x, y) + iv(x, y) with u(x, y) = e 2y cos(2x) and v(x, y) = e 2y sin(2x) u (a) To show f(z) is analytic, explicitly evaluate partials,,

More information

(1) Let f(z) be the principal branch of z 4i. (a) Find f(i). Solution. f(i) = exp(4i Log(i)) = exp(4i(π/2)) = e 2π. (b) Show that

(1) Let f(z) be the principal branch of z 4i. (a) Find f(i). Solution. f(i) = exp(4i Log(i)) = exp(4i(π/2)) = e 2π. (b) Show that Let fz be the principal branch of z 4i. a Find fi. Solution. fi = exp4i Logi = exp4iπ/2 = e 2π. b Show that fz fz 2 fz z 2 fz fz 2 = λfz z 2 for all z, z 2 0, where λ =, e 8π or e 8π. Proof. We have =

More information

Complex Function. Chapter Complex Number. Contents

Complex Function. Chapter Complex Number. Contents Chapter 6 Complex Function Contents 6. Complex Number 3 6.2 Elementary Functions 6.3 Function of Complex Variables, Limit and Derivatives 3 6.4 Analytic Functions and Their Derivatives 8 6.5 Line Integral

More information

Math 417 Midterm Exam Solutions Friday, July 9, 2010

Math 417 Midterm Exam Solutions Friday, July 9, 2010 Math 417 Midterm Exam Solutions Friday, July 9, 010 Solve any 4 of Problems 1 6 and 1 of Problems 7 8. Write your solutions in the booklet provided. If you attempt more than 5 problems, you must clearly

More information

Math Homework 2

Math Homework 2 Math 73 Homework Due: September 8, 6 Suppose that f is holomorphic in a region Ω, ie an open connected set Prove that in any of the following cases (a) R(f) is constant; (b) I(f) is constant; (c) f is

More information

MTH4101 CALCULUS II REVISION NOTES. 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) ax 2 + bx + c = 0. x = b ± b 2 4ac 2a. i = 1.

MTH4101 CALCULUS II REVISION NOTES. 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) ax 2 + bx + c = 0. x = b ± b 2 4ac 2a. i = 1. MTH4101 CALCULUS II REVISION NOTES 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) 1.1 Introduction Types of numbers (natural, integers, rationals, reals) The need to solve quadratic equations:

More information

MA3111S COMPLEX ANALYSIS I

MA3111S COMPLEX ANALYSIS I MA3111S COMPLEX ANALYSIS I 1. The Algebra of Complex Numbers A complex number is an expression of the form a + ib, where a and b are real numbers. a is called the real part of a + ib and b the imaginary

More information

6. Residue calculus. where C is any simple closed contour around z 0 and inside N ε.

6. Residue calculus. where C is any simple closed contour around z 0 and inside N ε. 6. Residue calculus Let z 0 be an isolated singularity of f(z), then there exists a certain deleted neighborhood N ε = {z : 0 < z z 0 < ε} such that f is analytic everywhere inside N ε. We define Res(f,

More information

Solutions to Selected Exercises. Complex Analysis with Applications by N. Asmar and L. Grafakos

Solutions to Selected Exercises. Complex Analysis with Applications by N. Asmar and L. Grafakos Solutions to Selected Exercises in Complex Analysis with Applications by N. Asmar and L. Grafakos Section. Complex Numbers Solutions to Exercises.. We have i + i. So a and b. 5. We have So a 3 and b 4.

More information

Syllabus: for Complex variables

Syllabus: for Complex variables EE-2020, Spring 2009 p. 1/42 Syllabus: for omplex variables 1. Midterm, (4/27). 2. Introduction to Numerical PDE (4/30): [Ref.num]. 3. omplex variables: [Textbook]h.13-h.18. omplex numbers and functions,

More information

Synopsis of Complex Analysis. Ryan D. Reece

Synopsis of Complex Analysis. Ryan D. Reece Synopsis of Complex Analysis Ryan D. Reece December 7, 2006 Chapter Complex Numbers. The Parts of a Complex Number A complex number, z, is an ordered pair of real numbers similar to the points in the real

More information

Solutions to Exercises 1.1

Solutions to Exercises 1.1 Section 1.1 Complex Numbers 1 Solutions to Exercises 1.1 1. We have So a 0 and b 1. 5. We have So a 3 and b 4. 9. We have i 0+ 1i. i +i because i i +i 1 {}}{ 4+4i + i 3+4i. 1 + i 3 7 i 1 3 3 + i 14 1 1

More information

18.04 Practice problems exam 1, Spring 2018 Solutions

18.04 Practice problems exam 1, Spring 2018 Solutions 8.4 Practice problems exam, Spring 8 Solutions Problem. omplex arithmetic (a) Find the real and imaginary part of z + z. (b) Solve z 4 i =. (c) Find all possible values of i. (d) Express cos(4x) in terms

More information

MTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106

MTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106 Name (Last name, First name): MTH 02 omplex Variables Final Exam May, 207 :0pm-5:0pm, Skurla Hall, Room 06 Exam Instructions: You have hour & 50 minutes to complete the exam. There are a total of problems.

More information

Considering our result for the sum and product of analytic functions, this means that for (a 0, a 1,..., a N ) C N+1, the polynomial.

Considering our result for the sum and product of analytic functions, this means that for (a 0, a 1,..., a N ) C N+1, the polynomial. Lecture 3 Usual complex functions MATH-GA 245.00 Complex Variables Polynomials. Construction f : z z is analytic on all of C since its real and imaginary parts satisfy the Cauchy-Riemann relations and

More information

Part IB. Complex Analysis. Year

Part IB. Complex Analysis. Year Part IB Complex Analysis Year 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2018 Paper 1, Section I 2A Complex Analysis or Complex Methods 7 (a) Show that w = log(z) is a conformal

More information

Math Spring 2014 Solutions to Assignment # 8 Completion Date: Friday May 30, 2014

Math Spring 2014 Solutions to Assignment # 8 Completion Date: Friday May 30, 2014 Math 3 - Spring 4 Solutions to Assignment # 8 ompletion Date: Friday May 3, 4 Question. [p 49, #] By finding an antiderivative, evaluate each of these integrals, where the path is any contour between the

More information

Exercises for Part 1

Exercises for Part 1 MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x + iy, x,y

More information

1 Discussion on multi-valued functions

1 Discussion on multi-valued functions Week 3 notes, Math 7651 1 Discussion on multi-valued functions Log function : Note that if z is written in its polar representation: z = r e iθ, where r = z and θ = arg z, then log z log r + i θ + 2inπ

More information

MATH 417 Homework 4 Instructor: D. Cabrera Due July 7. z c = e c log z (1 i) i = e i log(1 i) i log(1 i) = 4 + 2kπ + i ln ) cosz = eiz + e iz

MATH 417 Homework 4 Instructor: D. Cabrera Due July 7. z c = e c log z (1 i) i = e i log(1 i) i log(1 i) = 4 + 2kπ + i ln ) cosz = eiz + e iz MATH 47 Homework 4 Instructor: D. abrera Due July 7. Find all values of each expression below. a) i) i b) cos i) c) sin ) Solution: a) Here we use the formula z c = e c log z i) i = e i log i) The modulus

More information

Second Midterm Exam Name: Practice Problems March 10, 2015

Second Midterm Exam Name: Practice Problems March 10, 2015 Math 160 1. Treibergs Second Midterm Exam Name: Practice Problems March 10, 015 1. Determine the singular points of the function and state why the function is analytic everywhere else: z 1 fz) = z + 1)z

More information

EE2007 Tutorial 7 Complex Numbers, Complex Functions, Limits and Continuity

EE2007 Tutorial 7 Complex Numbers, Complex Functions, Limits and Continuity EE27 Tutorial 7 omplex Numbers, omplex Functions, Limits and ontinuity Exercise 1. These are elementary exercises designed as a self-test for you to determine if you if have the necessary pre-requisite

More information

Functions 45. Integrals, and Contours 55

Functions 45. Integrals, and Contours 55 MATH 43 COMPLEX ANALYSIS TRISTAN PHILLIPS These are notes from an introduction to complex analysis at the undergraduate level as taught by Paul Taylor at Shippensburg University during the Fall 26 semester.

More information

Conformal maps. Lent 2019 COMPLEX METHODS G. Taylor. A star means optional and not necessarily harder.

Conformal maps. Lent 2019 COMPLEX METHODS G. Taylor. A star means optional and not necessarily harder. Lent 29 COMPLEX METHODS G. Taylor A star means optional and not necessarily harder. Conformal maps. (i) Let f(z) = az + b, with ad bc. Where in C is f conformal? cz + d (ii) Let f(z) = z +. What are the

More information

Here are brief notes about topics covered in class on complex numbers, focusing on what is not covered in the textbook.

Here are brief notes about topics covered in class on complex numbers, focusing on what is not covered in the textbook. Phys374, Spring 2008, Prof. Ted Jacobson Department of Physics, University of Maryland Complex numbers version 5/21/08 Here are brief notes about topics covered in class on complex numbers, focusing on

More information

1 Res z k+1 (z c), 0 =

1 Res z k+1 (z c), 0 = 32. COMPLEX ANALYSIS FOR APPLICATIONS Mock Final examination. (Monday June 7..am 2.pm) You may consult your handwritten notes, the book by Gamelin, and the solutions and handouts provided during the Quarter.

More information

Qualifying Exam Complex Analysis (Math 530) January 2019

Qualifying Exam Complex Analysis (Math 530) January 2019 Qualifying Exam Complex Analysis (Math 53) January 219 1. Let D be a domain. A function f : D C is antiholomorphic if for every z D the limit f(z + h) f(z) lim h h exists. Write f(z) = f(x + iy) = u(x,

More information

2 Complex Functions and the Cauchy-Riemann Equations

2 Complex Functions and the Cauchy-Riemann Equations 2 Complex Functions and the Cauchy-Riemann Equations 2.1 Complex functions In one-variable calculus, we study functions f(x) of a real variable x. Likewise, in complex analysis, we study functions f(z)

More information

Math 411, Complex Analysis Definitions, Formulas and Theorems Winter y = sinα

Math 411, Complex Analysis Definitions, Formulas and Theorems Winter y = sinα Math 411, Complex Analysis Definitions, Formulas and Theorems Winter 014 Trigonometric Functions of Special Angles α, degrees α, radians sin α cos α tan α 0 0 0 1 0 30 π 6 45 π 4 1 3 1 3 1 y = sinα π 90,

More information

MTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106

MTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106 Name (Last name, First name): MTH 32 Complex Variables Final Exam May, 27 3:3pm-5:3pm, Skurla Hall, Room 6 Exam Instructions: You have hour & 5 minutes to complete the exam. There are a total of problems.

More information

Solutions to Complex Analysis Prelims Ben Strasser

Solutions to Complex Analysis Prelims Ben Strasser Solutions to Complex Analysis Prelims Ben Strasser In preparation for the complex analysis prelim, I typed up solutions to some old exams. This document includes complete solutions to both exams in 23,

More information

Notes on Complex Analysis

Notes on Complex Analysis Michael Papadimitrakis Notes on Complex Analysis Department of Mathematics University of Crete Contents The complex plane.. The complex plane...................................2 Argument and polar representation.........................

More information

The Residue Theorem. Integration Methods over Closed Curves for Functions with Singularities

The Residue Theorem. Integration Methods over Closed Curves for Functions with Singularities The Residue Theorem Integration Methods over losed urves for Functions with Singularities We have shown that if f(z) is analytic inside and on a closed curve, then f(z)dz = 0. We have also seen examples

More information

MATH243 First Semester 2013/14. Exercises 1

MATH243 First Semester 2013/14. Exercises 1 Complex Functions Dr Anna Pratoussevitch MATH43 First Semester 013/14 Exercises 1 Submit your solutions to questions marked with [HW] in the lecture on Monday 30/09/013 Questions or parts of questions

More information

Exercises for Part 1

Exercises for Part 1 MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x+iy, x,y R:

More information

Math 423/823 Exam 1 Topics Covered

Math 423/823 Exam 1 Topics Covered Math 423/823 Exam 1 Topics Covered Complex numbers: C: z = x+yi, where i 2 = 1; addition and multiplication behaves like reals. Formally, x + yi (x,y), with (x,y) + (a,b) = (x + a,y + b) and (x,y)(a,b)

More information

Residues and Contour Integration Problems

Residues and Contour Integration Problems Residues and ontour Integration Problems lassify the singularity of fz at the indicated point.. fz = cotz at z =. Ans. Simple pole. Solution. The test for a simple pole at z = is that lim z z cotz exists

More information

Solutions for Problem Set #5 due October 17, 2003 Dustin Cartwright and Dylan Thurston

Solutions for Problem Set #5 due October 17, 2003 Dustin Cartwright and Dylan Thurston Solutions for Problem Set #5 due October 17, 23 Dustin Cartwright and Dylan Thurston 1 (B&N 6.5) Suppose an analytic function f agrees with tan x, x 1. Show that f(z) = i has no solution. Could f be entire?

More information

Functions of a Complex Variable and Integral Transforms

Functions of a Complex Variable and Integral Transforms Functions of a Complex Variable and Integral Transforms Department of Mathematics Zhou Lingjun Textbook Functions of Complex Analysis with Applications to Engineering and Science, 3rd Edition. A. D. Snider

More information

Complex varibles:contour integration examples

Complex varibles:contour integration examples omple varibles:ontour integration eamples 1 Problem 1 onsider the problem d 2 + 1 If we take the substitution = tan θ then d = sec 2 θdθ, which leads to dθ = π sec 2 θ tan 2 θ + 1 dθ Net we consider the

More information

f (n) (z 0 ) Theorem [Morera s Theorem] Suppose f is continuous on a domain U, and satisfies that for any closed curve γ in U, γ

f (n) (z 0 ) Theorem [Morera s Theorem] Suppose f is continuous on a domain U, and satisfies that for any closed curve γ in U, γ Remarks. 1. So far we have seen that holomorphic is equivalent to analytic. Thus, if f is complex differentiable in an open set, then it is infinitely many times complex differentiable in that set. This

More information

Chapter 9. Analytic Continuation. 9.1 Analytic Continuation. For every complex problem, there is a solution that is simple, neat, and wrong.

Chapter 9. Analytic Continuation. 9.1 Analytic Continuation. For every complex problem, there is a solution that is simple, neat, and wrong. Chapter 9 Analytic Continuation For every complex problem, there is a solution that is simple, neat, and wrong. - H. L. Mencken 9.1 Analytic Continuation Suppose there is a function, f 1 (z) that is analytic

More information

Final Exam - MATH 630: Solutions

Final Exam - MATH 630: Solutions Final Exam - MATH 630: Solutions Problem. Find all x R satisfying e xeix e ix. Solution. Comparing the moduli of both parts, we obtain e x cos x, and therefore, x cos x 0, which is possible only if x 0

More information

Complex Analysis MATH 6300 Fall 2013 Homework 4

Complex Analysis MATH 6300 Fall 2013 Homework 4 Complex Analysis MATH 6300 Fall 2013 Homework 4 Due Wednesday, December 11 at 5 PM Note that to get full credit on any problem in this class, you must solve the problems in an efficient and elegant manner,

More information

Complex Variables Notes for Math 703. Updated Fall Anton R. Schep

Complex Variables Notes for Math 703. Updated Fall Anton R. Schep Complex Variables Notes for Math 703. Updated Fall 20 Anton R. Schep CHAPTER Holomorphic (or Analytic) Functions. Definitions and elementary properties In complex analysis we study functions f : S C,

More information

Assignment 2 - Complex Analysis

Assignment 2 - Complex Analysis Assignment 2 - Complex Analysis MATH 440/508 M.P. Lamoureux Sketch of solutions. Γ z dz = Γ (x iy)(dx + idy) = (xdx + ydy) + i Γ Γ ( ydx + xdy) = (/2)(x 2 + y 2 ) endpoints + i [( / y) y ( / x)x]dxdy interiorγ

More information

Complex Analysis. Travis Dirle. December 4, 2016

Complex Analysis. Travis Dirle. December 4, 2016 Complex Analysis 2 Complex Analysis Travis Dirle December 4, 2016 2 Contents 1 Complex Numbers and Functions 1 2 Power Series 3 3 Analytic Functions 7 4 Logarithms and Branches 13 5 Complex Integration

More information

1 z n = 1. 9.(Problem) Evaluate each of the following, that is, express each in standard Cartesian form x + iy. (2 i) 3. ( 1 + i. 2 i.

1 z n = 1. 9.(Problem) Evaluate each of the following, that is, express each in standard Cartesian form x + iy. (2 i) 3. ( 1 + i. 2 i. . 5(b). (Problem) Show that z n = z n and z n = z n for n =,,... (b) Use polar form, i.e. let z = re iθ, then z n = r n = z n. Note e iθ = cos θ + i sin θ =. 9.(Problem) Evaluate each of the following,

More information

(x 1, y 1 ) = (x 2, y 2 ) if and only if x 1 = x 2 and y 1 = y 2.

(x 1, y 1 ) = (x 2, y 2 ) if and only if x 1 = x 2 and y 1 = y 2. 1. Complex numbers A complex number z is defined as an ordered pair z = (x, y), where x and y are a pair of real numbers. In usual notation, we write z = x + iy, where i is a symbol. The operations of

More information

Complex Variables...Review Problems (Residue Calculus Comments)...Fall Initial Draft

Complex Variables...Review Problems (Residue Calculus Comments)...Fall Initial Draft Complex Variables........Review Problems Residue Calculus Comments)........Fall 22 Initial Draft ) Show that the singular point of fz) is a pole; determine its order m and its residue B: a) e 2z )/z 4,

More information

MA 412 Complex Analysis Final Exam

MA 412 Complex Analysis Final Exam MA 4 Complex Analysis Final Exam Summer II Session, August 9, 00.. Find all the values of ( 8i) /3. Sketch the solutions. Answer: We start by writing 8i in polar form and then we ll compute the cubic root:

More information

1 Complex Numbers. 1.1 Sums and Products

1 Complex Numbers. 1.1 Sums and Products 1 Complex Numbers 1.1 Sums Products Definition: The complex plane, denoted C is the set of all ordered pairs (x, y) with x, y R, where Re z = x is called the real part Imz = y is called the imaginary part.

More information

III. Consequences of Cauchy s Theorem

III. Consequences of Cauchy s Theorem MTH6 Complex Analysis 2009-0 Lecture Notes c Shaun Bullett 2009 III. Consequences of Cauchy s Theorem. Cauchy s formulae. Cauchy s Integral Formula Let f be holomorphic on and everywhere inside a simple

More information

Topic 2 Notes Jeremy Orloff

Topic 2 Notes Jeremy Orloff Topic 2 Notes Jeremy Orloff 2 Analytic functions 2.1 Introduction The main goal of this topic is to define and give some of the important properties of complex analytic functions. A function f(z) is analytic

More information

MATH 311: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE

MATH 311: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE Recall the Residue Theorem: Let be a simple closed loop, traversed counterclockwise. Let f be a function that is analytic on and meromorphic inside. Then

More information

Math 185 Homework Exercises II

Math 185 Homework Exercises II Math 185 Homework Exercises II Instructor: Andrés E. Caicedo Due: July 10, 2002 1. Verify that if f H(Ω) C 2 (Ω) is never zero, then ln f is harmonic in Ω. 2. Let f = u+iv H(Ω) C 2 (Ω). Let p 2 be an integer.

More information

Math 421 Midterm 2 review questions

Math 421 Midterm 2 review questions Math 42 Midterm 2 review questions Paul Hacking November 7, 205 () Let U be an open set and f : U a continuous function. Let be a smooth curve contained in U, with endpoints α and β, oriented from α to

More information

MAT389 Fall 2016, Problem Set 11

MAT389 Fall 2016, Problem Set 11 MAT389 Fall 216, Problem Set 11 Improper integrals 11.1 In each of the following cases, establish the convergence of the given integral and calculate its value. i) x 2 x 2 + 1) 2 ii) x x 2 + 1)x 2 + 2x

More information

Theorem [Mean Value Theorem for Harmonic Functions] Let u be harmonic on D(z 0, R). Then for any r (0, R), u(z 0 ) = 1 z z 0 r

Theorem [Mean Value Theorem for Harmonic Functions] Let u be harmonic on D(z 0, R). Then for any r (0, R), u(z 0 ) = 1 z z 0 r 2. A harmonic conjugate always exists locally: if u is a harmonic function in an open set U, then for any disk D(z 0, r) U, there is f, which is analytic in D(z 0, r) and satisfies that Re f u. Since such

More information

Exercises involving elementary functions

Exercises involving elementary functions 017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 1 Exercises involving elementary functions 1. This question was in the class test in 016/7 and was worth 5 marks. a) Let

More information

u = 0; thus v = 0 (and v = 0). Consequently,

u = 0; thus v = 0 (and v = 0). Consequently, MAT40 - MANDATORY ASSIGNMENT #, FALL 00; FASIT REMINDER: The assignment must be handed in before 4:30 on Thursday October 8 at the Department of Mathematics, in the 7th floor of Niels Henrik Abels hus,

More information

Solutions to Tutorial for Week 3

Solutions to Tutorial for Week 3 The University of Sydney School of Mathematics and Statistics Solutions to Tutorial for Week 3 MATH9/93: Calculus of One Variable (Advanced) Semester, 08 Web Page: sydney.edu.au/science/maths/u/ug/jm/math9/

More information

Introduction to Complex Analysis

Introduction to Complex Analysis Introduction to Complex Analysis George Voutsadakis 1 1 Mathematics and Computer Science Lake Superior State University LSSU Math 413 George Voutsadakis (LSSU) Complex Analysis October 2014 1 / 67 Outline

More information

Physics 307. Mathematical Physics. Luis Anchordoqui. Wednesday, August 31, 16

Physics 307. Mathematical Physics. Luis Anchordoqui. Wednesday, August 31, 16 Physics 307 Mathematical Physics Luis Anchordoqui 1 Bibliography L. A. Anchordoqui and T. C. Paul, ``Mathematical Models of Physics Problems (Nova Publishers, 2013) G. F. D. Duff and D. Naylor, ``Differential

More information

Complex numbers, the exponential function, and factorization over C

Complex numbers, the exponential function, and factorization over C Complex numbers, the exponential function, and factorization over C 1 Complex Numbers Recall that for every non-zero real number x, its square x 2 = x x is always positive. Consequently, R does not contain

More information

Chapter 30 MSMYP1 Further Complex Variable Theory

Chapter 30 MSMYP1 Further Complex Variable Theory Chapter 30 MSMYP Further Complex Variable Theory (30.) Multifunctions A multifunction is a function that may take many values at the same point. Clearly such functions are problematic for an analytic study,

More information

Solutions to Exercises 6.1

Solutions to Exercises 6.1 34 Chapter 6 Conformal Mappings Solutions to Exercises 6.. An analytic function fz is conformal where f z. If fz = z + e z, then f z =e z z + z. We have f z = z z += z =. Thus f is conformal at all z.

More information

The Calculus of Residues

The Calculus of Residues hapter 7 The alculus of Residues If fz) has a pole of order m at z = z, it can be written as Eq. 6.7), or fz) = φz) = a z z ) + a z z ) +... + a m z z ) m, 7.) where φz) is analytic in the neighborhood

More information

Part IB. Further Analysis. Year

Part IB. Further Analysis. Year Year 2004 2003 2002 2001 10 2004 2/I/4E Let τ be the topology on N consisting of the empty set and all sets X N such that N \ X is finite. Let σ be the usual topology on R, and let ρ be the topology on

More information

Evaluation of integrals

Evaluation of integrals Evaluation of certain contour integrals: Type I Type I: Integrals of the form 2π F (cos θ, sin θ) dθ If we take z = e iθ, then cos θ = 1 (z + 1 ), sin θ = 1 (z 1 dz ) and dθ = 2 z 2i z iz. Substituting

More information

Selected Solutions To Problems in Complex Analysis

Selected Solutions To Problems in Complex Analysis Selected Solutions To Problems in Complex Analysis E. Chernysh November 3, 6 Contents Page 8 Problem................................... Problem 4................................... Problem 5...................................

More information

Integration in the Complex Plane (Zill & Wright Chapter 18)

Integration in the Complex Plane (Zill & Wright Chapter 18) Integration in the omplex Plane Zill & Wright hapter 18) 116-4-: omplex Variables Fall 11 ontents 1 ontour Integrals 1.1 Definition and Properties............................. 1. Evaluation.....................................

More information

Math Spring 2014 Solutions to Assignment # 12 Completion Date: Thursday June 12, 2014

Math Spring 2014 Solutions to Assignment # 12 Completion Date: Thursday June 12, 2014 Math 3 - Spring 4 Solutions to Assignment # Completion Date: Thursday June, 4 Question. [p 67, #] Use residues to evaluate the improper integral x + ). Ans: π/4. Solution: Let fz) = below. + z ), and for

More information

EE2007: Engineering Mathematics II Complex Analysis

EE2007: Engineering Mathematics II Complex Analysis EE2007: Engineering Mathematics II omplex Analysis Ling KV School of EEE, NTU ekvling@ntu.edu.sg V4.2: Ling KV, August 6, 2006 V4.1: Ling KV, Jul 2005 EE2007 V4.0: Ling KV, Jan 2005, EE2007 V3.1: Ling

More information

MTH3101 Spring 2017 HW Assignment 4: Sec. 26: #6,7; Sec. 33: #5,7; Sec. 38: #8; Sec. 40: #2 The due date for this assignment is 2/23/17.

MTH3101 Spring 2017 HW Assignment 4: Sec. 26: #6,7; Sec. 33: #5,7; Sec. 38: #8; Sec. 40: #2 The due date for this assignment is 2/23/17. MTH0 Spring 07 HW Assignment : Sec. 6: #6,7; Sec. : #5,7; Sec. 8: #8; Sec. 0: # The due date for this assignment is //7. Sec. 6: #6. Use results in Sec. to verify that the function g z = ln r + iθ r >

More information

MATH FINAL SOLUTION

MATH FINAL SOLUTION MATH 185-4 FINAL SOLUTION 1. (8 points) Determine whether the following statements are true of false, no justification is required. (1) (1 point) Let D be a domain and let u,v : D R be two harmonic functions,

More information

Complex Analysis Important Concepts

Complex Analysis Important Concepts Complex Analysis Important Concepts Travis Askham April 1, 2012 Contents 1 Complex Differentiation 2 1.1 Definition and Characterization.............................. 2 1.2 Examples..........................................

More information

Topic 7 Notes Jeremy Orloff

Topic 7 Notes Jeremy Orloff Topic 7 Notes Jeremy Orloff 7 Taylor and Laurent series 7. Introduction We originally defined an analytic function as one where the derivative, defined as a limit of ratios, existed. We went on to prove

More information

MORE CONSEQUENCES OF CAUCHY S THEOREM

MORE CONSEQUENCES OF CAUCHY S THEOREM MOE CONSEQUENCES OF CAUCHY S THEOEM Contents. The Mean Value Property and the Maximum-Modulus Principle 2. Morera s Theorem and some applications 3 3. The Schwarz eflection Principle 6 We have stated Cauchy

More information

Class test: week 10, 75 minutes. (30%) Final exam: April/May exam period, 3 hours (70%).

Class test: week 10, 75 minutes. (30%) Final exam: April/May exam period, 3 hours (70%). 17-4-2013 12:55 c M. K. Warby MA3914 Complex variable methods and applications 0 1 MA3914 Complex variable methods and applications Lecture Notes by M.K. Warby in 2012/3 Department of Mathematical Sciences

More information

1. COMPLEX NUMBERS. z 1 + z 2 := (a 1 + a 2 ) + i(b 1 + b 2 ); Multiplication by;

1. COMPLEX NUMBERS. z 1 + z 2 := (a 1 + a 2 ) + i(b 1 + b 2 ); Multiplication by; 1. COMPLEX NUMBERS Notations: N the set of the natural numbers, Z the set of the integers, R the set of real numbers, Q := the set of the rational numbers. Given a quadratic equation ax 2 + bx + c = 0,

More information

INTEGRATION WORKSHOP 2004 COMPLEX ANALYSIS EXERCISES

INTEGRATION WORKSHOP 2004 COMPLEX ANALYSIS EXERCISES INTEGRATION WORKSHOP 2004 COMPLEX ANALYSIS EXERCISES PHILIP FOTH 1. Cauchy s Formula and Cauchy s Theorem 1. Suppose that γ is a piecewise smooth positively ( counterclockwise ) oriented simple closed

More information

Complex Analysis Homework 1: Solutions

Complex Analysis Homework 1: Solutions Complex Analysis Fall 007 Homework 1: Solutions 1.1.. a) + i)4 + i) 8 ) + 1 + )i 5 + 14i b) 8 + 6i) 64 6) + 48 + 48)i 8 + 96i c) 1 + ) 1 + i 1 + 1 i) 1 + i)1 i) 1 + i ) 5 ) i 5 4 9 ) + 4 4 15 i ) 15 4

More information

1. The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions.

1. The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions. Complex Analysis Qualifying Examination 1 The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions 2 ANALYTIC FUNCTIONS:

More information

MTH 3102 Complex Variables Solutions: Practice Exam 2 Mar. 26, 2017

MTH 3102 Complex Variables Solutions: Practice Exam 2 Mar. 26, 2017 Name Last name, First name): MTH 31 omplex Variables Solutions: Practice Exam Mar. 6, 17 Exam Instructions: You have 1 hour & 1 minutes to complete the exam. There are a total of 7 problems. You must show

More information

MAT665:ANALYTIC FUNCTION THEORY

MAT665:ANALYTIC FUNCTION THEORY MAT665:ANALYTIC FUNCTION THEORY DR. RITU AGARWAL MALAVIYA NATIONAL INSTITUTE OF TECHNOLOGY JAIPUR Contents 1. About 2 2. Complex Numbers 2 3. Fundamental inequalities 2 4. Continuously differentiable functions

More information

Suggested Homework Solutions

Suggested Homework Solutions Suggested Homework Solutions Chapter Fourteen Section #9: Real and Imaginary parts of /z: z = x + iy = x + iy x iy ( ) x iy = x #9: Real and Imaginary parts of ln z: + i ( y ) ln z = ln(re iθ ) = ln r

More information

Types of Real Integrals

Types of Real Integrals Math B: Complex Variables Types of Real Integrals p(x) I. Integrals of the form P.V. dx where p(x) and q(x) are polynomials and q(x) q(x) has no eros (for < x < ) and evaluate its integral along the fol-

More information

Taylor and Laurent Series

Taylor and Laurent Series Chapter 4 Taylor and Laurent Series 4.. Taylor Series 4... Taylor Series for Holomorphic Functions. In Real Analysis, the Taylor series of a given function f : R R is given by: f (x + f (x (x x + f (x

More information

lim when the limit on the right exists, the improper integral is said to converge to that limit.

lim when the limit on the right exists, the improper integral is said to converge to that limit. hapter 7 Applications of esidues - evaluation of definite and improper integrals occurring in real analysis and applied math - finding inverse Laplace transform by the methods of summing residues. 6. Evaluation

More information