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.

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1 . 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, that is, express each in standard Cartesian form x + iy. (a) (e) (g) (h) ( i) 3 ( ) 3 + i i I( + i) 3 ( R ) 3 + i (a) ( i) 3 = i. (e) ( + i) 3 6 = i i. (g) Note + i = e πi/4 and ( + i) 3 = 3/ e 3πi/4 so I( + i) 3 = 3/ sin (3π/4) =. (h) /8..3.(Problem) With a labeled sketch, show the point sets defined by the following. (a) z 4 (c) z + i < (i) z + = z + (a) z 4 can be read as a set of points inside a circle with a radius of 4 centered at z =. (c) Read it as z ( + i) <, i.e. a set of points inside the circle with a radius centered at + i. (i) Square the both sides to get z + = z + z + which is expanded to (x + ) + y = x + y + x + y + from which one gets x = x + y, i.e. y = 0 and x > 0 (don t forget this condition!)..(problem) Determine the range R for the given function. Include sketches of both the domain D and the range R, and give the equations of any curved parts of the boundary of R. (a) w(z) = z + + i on 0 < x <, 0 < y < (d) w(z) = z on < x < 0, 0 < y <

2 (f) w(z) = iz on 0 < x <, 0 < y < (g) w(z) = z 3 on 0 < x <, 0 < y < (a) Shift the rectangle (0 < x <, 0 < y < ) to the right by and up by. y v w = z++i x 3 (d) Note that with w = z, the argument on z is doubled. So the positive y axis (θ = π/) is mapped to the negative v axis (ϕ = π) and the negative x axis (θ = π) is mapped to the positive u axis (ϕ = π), i.e. the area expressed by v 0. y v u w = z u x (f) w = iz is equivalent to u + iv = i(x + iy). Comparing both the real part and the imaginary part gives u = xy and v = x y. Using this relationship, one can transform the boundary of the rectangle (0 < x <, 0 < y < ). i.e. { x = 0, 0 < y < } {u = 0, < v < 0}, { y = 0, 0 < x < } {u = 0, 0 < v < }, { x =, 0 < y < } {u = y, v = y } {v = u /4}, {y =, 0 < x < } {u = y, v = y } {v = u /4 }.

3 y v w = i z v = - u /4 x - v = - + u /4 u (g) Note that with w = z 3, the argument is tripled. So the positive x axis (θ = 0) remains the positive u axis while the positive y axis (θ = π/) is mapped to the negative v axis (θ = 3π/). So the first quadrant is tripled on the w plane. y w = z 3 v x u 3.(Problem) Show whether or not e z = e z. More generally, is w (z) = w ( z )? Thus, On the other hand, so e z = e (x+iy) = e x e iy = e x (cos y + i sin y). e z = e x cos y + i sin y = e x. e z = e x +y, e z = e z. Counterexample: z = iy. e z = cos y + i sin y = while e z = e y. In general, w(z) = w( z ). 3

4 4.(Problem) Show whether or not More generally, is Yes as and ē z = e z. w(z) = w ( z)? e z = e x (cos y + i sin y) = e x (cos y i sin y), e z = e x iy = e x (cos y i sin y). This is not true in general. Counterexample: w(z) = i. 7.(Problem) Show that (a) sin ( z) = sin z and cos ( z) = cos z (b) cos (z + z ) = cos z cos z sin z sin z (e) cos (x + iy) = cos x cosh y i sin x sinh y (f) sin (x + iy) = sin x cosh y + i cos x sinh y (g) cosh z sinh z = (a)(b) Both are proven by analytical continuation. equations (4) and (5). If you are not satisfied by this, use the definition of (e) Note that cos (iy) = cos (x + iy) = cos x cos iy sin x sin iy, ( e i(iy) + e i(iy)) = (e y + e y ) = cosh y, and so sin (iy) = i ( e i(iy) e i(iy)) = i ( e y e y) = i sinh y, cos (x + iy) = cos x cosh y i sin x sinh y. (f) sin (x + iy) = sin x cos iy + cos x sin iy = sin x cosh y + i cos x sinh y. (g) cosh z sinh z = ( e z + e z ) ( e z e z ) () = ez + + e z ez + e z () 4 4 =. (3) 8.(Problem) Show that (b) cosh (z + z ) = cosh z cosh z + sinh z sinh z. 4

5 (g) Solution (b) Expand cosh z sinh z =. e (z+z) + e (z+z) and e z + e z e z + e z + ez e z separately and verify that the both are the same. (g) (e z + e z ) (e z e z e z e z ) e z + + e z (e z + e z ) = 4 9. (Problem) Evaluate each of the following in standard Cartesian form. (a) e +πi (d) sin (3 + πi) (a) e e πi = e (cos π + i sin π) = e. (d).prove that sin (3 + πi) = ei(3+πi) e i(3+πi) i = e3i e π e 3i e π i =... (a) e z = if and only if z = nπi, where n is any integer. = cosh π sin 3 + i cos 3 sinh π. (b) e z = e z if and only if z = z + nπi, where n is any integer. (a) This comes from = e nπi (polar form). (b) e z = e z is equivalent to e z z =. From the result of (a), z z must be equal to nπ. 3.(Problem) Show that the range R of the function sin z on the semi-infinite strip π/ < x < π/, 0 < y < is shown in the accompanying figure. sin z = sin (x + iy) = sin x cosh y + i cos x sin hy, i.e. u = sin x cosh y and v = cos x sin hy. The bottom (y = 0, π/ < x < π/) is mapped to u = sin x, v = 0 or < u <, v = 0. The right edge (x = π/, y > 0) is mapped to u = cosh y, v = 0 or < u <, v = 0. The left edge (x = π/, y > 0) is mapped to u = cosh y, v = 0 or < u <, v = 0. Combining them, it is found that the rectangle is mapped to the upper half plane. =. 5

6 .4.(Problem) Determine r and the principal argument θ 0 (in radians and in degrees) for each of the following values of z. (a) 3i (f) i (a) 3i = 3e 3/πi. (f) i = 37e.405i. 4.(Problem) Obtain z 0 and z 0, in both polar and Cartesian form, for each given z. (a) + i (a) z = e 3π/4i. z 0 = 5 e 7.5πi = 3i. z 0 = 0 e 5πi = (Problem) Find all values of z / and z /5 for each given z. Express those values in polar form, and show their location in the z plane, as we have done in Fig. 5. (a) i (a) z = e π/i so z / = e π/4i and z /5 = e π/0i. 6.(Problem) Obtain, in Cartesian form, all values of log z for each given z. (a) (a) = e πi+nπi so ln ( ) = ln + (n + )πi. 8.(Problem) Obtain, in polar form, all values of z /3, z 3/, and z π for each given z. (a) i (a) z = e (π/+nπ)i so z /3 = /3 e /3(π/+nπ)i = /3 e π/3i e 4n/3πi..(Problem) Obtain, in Cartesian form, the principal values of log z and z for each given z. (a) 3i 6

7 (a) ln ( 3i) = ln ( 3e (3π/+nπ)i) = ln 3 + i(3π/ + nπ). By setting n = 0, the principal value of ln ( 3) is ln 3 + 3πi/. 3i = (3e (3π/+nπ)i ) = 3e (3π/+nπ)i/. By setting n = 0, the principal value of 3i is 3e 3πi/4 = 3 ( + i ). 3.(Problem) (Inverse of sine function) We define the inverse of the sine function such that z = sin w (a) Writing the latter as w(z) = arcsin z (4) z = ( e iw e iw) /i, (5) show that e iw = iz + ( z ) /, and hence that arcsin z = i log [iz + ] z. (6) (b) Observe that arcsin z is multi-valued because of the ( z ) / and also because of the log [ ]. Specifically, for each value of z ( ±), the ( z ) / gives two values. Then, for each of these values the log gives an infinite set of values. To illustrate this point, show that for k = 0, ±, ±,... arcsin = π 5π + kπ or kπ (7) (c) Determine all possible values of arcsin. (d) Determine all possible values of arcsin (i). (a) z = (e iw e iw )/(i) is equivalent to iz = e iw e iw or (e iw ) ize iw = 0. Solving this quadratic equation, one gets e iw = iz + ( z ) /, i.e. w = arcsin z = i ln(iz + ( z ) / ). Note that ( z ) / is multi-valued (±) without an appropriate branch cut which should take care of the ± sign when solving the quadratic equation. (b) Note that z / = (re θi+nπi/ ) = re θ i+nπi where the values of e nπi is either or - depending on whether n is even or odd. Therefore, ( (/) ) / = ± 3. arcsin (/) = i ln(i/ ± 3/) = i ln(e π/6i+kπi )or i ln(e 5π/6i+kπi ) = π 6 + kπor 6π 6 + kπ. (c) arcsin = i ln(i + 3) = i ln(( ± 3)i) = i ln(( ± 3)e (π/+kπ)i ) = i(ln( ± 3) + i(π/ + kπ)) = (π/ + kπ) i ln ( ± 3). (d) arcsin (i) = i ln(i(i)+( (i) ) / ) = i ln( ± 5) = nπ i ln ( 5 )or (n+)π i ln( 5+). 4. (Problem) (Inverse of other trigonometric functions) Proceeding as in Exercise 3, derive these formulas. (a) arccos z = i log [ z + z ] (b) arctan z = i (c) cot z = i log i z i+z log z+ z 7

8 (a) Solve z = eiw +e iw for e iw to get arccos z = i ln(z + z / ). (b) Solve z = (c) Similar to (b). eiw e iw i(e iw +e iw ) to get arctanz = i i z ln i+z..5 0.(Problem) Given f (z), use (9) to obtain f (z). Express your answer in terms of z. (a) cos z (e) z (z ) (a) f (z) = sin z. (e) f (z) = /z. If f(z) is a regular function, you can differentiate it as if it were a real valued function..(problem) Given f (z), determine f (z), where it exists, and state where f is analytic and where it is not. (a) ( z 3 ) 5 (b) (c) (d) (e) (f) x + iy x + y z sin z z + 3iz z 3 + x + i sin y (a) f(z) is regular so f (z) = 5( z 3 ) 4 ( 6z ). (b) Note that which is NOT analytic (contains z). (c) This is not analytic as it contains x + iy x + y = z z z = z z = z z. 8

9 (d) f (z) = (z + 3i) ( + 3iz + z ). (e) f (z) = 3z ( + z 3 ). (f) This is not an analytic function (use C-R to verify)..(problem) (Cauchy-Riemann equations in polar coordinates) Derive the Cauchy-Riemann equations (30) in the manner indicated. (a) By carrying out the limit in (8). HINT: First let z 0 along the constant-θ line through z 0, and then let z 0 along the constant-r line through z 0. Pay careful attention to your expression for z in each of these cases because these cases are trickier than the cases of a horizontal approach ( z = x) and a vertical approach ( z = i y), used in (6) and (7). (b) By making the change of variables x = r cos θ, y = r sin θ in (8). (a) On the constant θ line, z = re iθ (r varies but θ remains constant) so f (z) = u(r + r, θ) + iv(r + r, θ) u(r, θ) iv(r, θ) lim r 0 re iθ (8) = lim u(r + r, θ) u(r) v(r + r, θ) v(r) ) + i r 0 r r (9) = e iθ (u r + iv r ). (0) On the constant-r line, z = i r θ e iθ (r is constant while θ varies) so f u(r, θ + θ) + iv(r, θ + θ) u(r, θ) iv(r, θ) (z) = lim θ 0 i r θ e iθ () By comparing the real and imaginary parts, one obtains = r e iθ v θ i r e iθ u θ. () (b) Recall the chain differentiation rule, i.e. u r = r v θ v r = r u θ. (3) u x = u r r x + u θ θ x (4) v y = v r r y + v θ θ y, (5) and 9

10 So u x = v y u r r x + u θ θ x = v r r y + v θ θ x or r x = r x = x + y x x = x + y = r cos θ r = cos θ etc... sin θ u r cos θ u θ r and u y = v x u r r y + u θ θ y = v r r x v θ θ y or cos θ = v r sin θ + v θ. r u r sin θ + u θ cos θ r sin θ = v r cos θ + v θ. r Solving the above simultaneous equations for u r and v r yields u r = r v θ and v r = r u θ. 3.(Problem) Determine where these functions are differentiable and where they are analytic, by checking for satisfaction of the Cauchy-Riemann equations and for continuity of u, v and their first-order partial derivatives. (a) f (z) = z 00 (b) f (z) = z, defined by the branch cut shown in Fig. 6 (c) f (z) = z, where z is defined by the branch cut shown in Fig.6 (a) f(z) = z 00 is analytic as it is a function of z alone. f (z) = 00z 99. (b) f(z) = z is analytic except for z = 0 at which point f (z) fails to exist. f (z) = z. (c) f(z) = z is analytic except for z = 0 at which point f (z) fails to exist. f (z) = z 3/. 4.(Problem) (a) f = u + iv and f = u iv. The Cauchy-Riemann relation requires that u x = v y, u y = v x, u x = v y and u y = v x from which u x = u y = 0 so u = const then, v x = v y = 0 so v = const too. (b) By the fundamental theorem of calculus. 5. Determine whether or not the given function u is harmonic and, if so, in what region. If it is, find the most general conjugate function v and corresponding analytic function f (z). Express f in terms of z. 0

11 (a) e x cos y (b) e x sin y (c) x 3 3xy (d) r 3 sin 3θ (e) r cos θ + 4 (f) r (g) x cos x cosh y + y sin x sinh y (In the following solution, add a constant C to f(z).) (a) Yes. v = e x sin y so that f(z) = e z. (b) Yes. v = e x cos y. Note that (c) Yes. v = 3x y y 3 so f(z) = z 3. u + iv = e x (sin y i cos y) = e x ( i sin y i cos y) = ie x (cos y + i sin y) = ie x e yi = ie (x+iy) = ie z. (d) Yes. Take v = r 3 cos 3θ so that u+iv = r 3 (sin 3θ i cos 3θ) = r 3 ( i sin 3θ i cos 3θ) = r 3 ( i)(cos 3θ+ i sin 3θ) = i(r 3 e 3θi ) = iz 3. (e) Yes. v = r sin θ so that f(z) = z + 4. (f) No. (g) Yes. v = y cos x cosh y x sin x sin hy.