Math Homework 1. The homework consists mostly of a selection of problems from the suggested books. 1 ± i ) 2 = 1, 2.

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1 Math Homework 1 September 1, 006 The homework consists mostly of a selection of problems from the suggested books. 1. (a) Find the value of (1 + i) n + (1 i) n for every n N. We will use the polar form of 1 + i = (cos(π/4) + i sin(π/4)) and De Moivre s formula. Since 1 + i = 1 i, (1 + i) n = (1 + i) n = (1 i) n we have (1 + i) n + (1 i) n = R(1 + i) n = R n/ (cos(nπ/4) + i sin(nπ/4)) = n/ cos(nπ/4) = n/+1 cos(nπ/4). (b) Show that ( 1 ± i ) 3 3 = 1, ( ±1 ± i ) 6 3 = 1. Since x 3 1 = (x 1)(x + x + 1) it suffices to prove that satisfies ρ 1, + ρ 1, + 1 = 0. We have ( ρ 1, = 1 ± i 3 1 ± i ) 3 = 1 3 i 3 4 From this ρ 1, + ρ 1, + 1 = 0 follows easily. = 1 i 3. For ±1 ± i 3 we argue as follows: if the signs are opposite we are considering ±ρ 1,. Since ρ 3 1, = 1, ρ 6 1, = 1. For the other choices r 1 = 1 + i 3, r = 1 i 3, r = r 1 which are called the sixth primitive roots of unity, we argue as follows: x 6 1 = (x 3 1)(x + 1)(x x + 1) implies that we should prove that r1 r = 0. We calculate r1 = i 3 = 1 + i 3 4 1

2 from which the result follows. This way we avoid calculating sixth powers. (c) Find the fourth roots of 1. We write 1 = 1(cos π + i sin π) (polar form) and apply De Moivre s formulas. The fourth roots are: r k = 4 ( ( ) ( )) π + πk π + πk 1 cos + i sin, k = 0, 1,, This gives r 0 = cos(π/4)+i sin(π/4) = +i, r 1 = cos(3π/4)+i sin(3π/4) = +i r = cos(5π/4)+i sin(5π/4) = i, r 3 = cos(7π/4)+i sin(7π/4) = i. Find the conditions under which the equation az +b z +c = 0 has exactly one solution and compute the solution. We write the equation and its conjugate as a system az + b z + c = 0 bz + ā z + c = 0. We think of the variables z and z as independent variables. This system has exactly one solution when its determinant is nonzero, i.e., a b b = a b 0. ā In this case the solution is given by Cramer s rule as c b c ā z = a b = cb cā a b. We can also solve for z and we see that indeed it is conjugate of z: a c b c z = a b = bc a c a b. Alternatively we set the real and imaginary parts of the equation to be 0. We set a = α 1 + iα, b = β 1 + iβ, c = γ 1 + iγ, z = x + iy to get α 1 x α y + β 1 x + β y + γ 1 = 0, α x + α 1 y + β x β 1 y + γ = 0,.

3 which gives the system (α 1 + β 1 )x + ( α + β )y + γ 1 = 0 (α + β )x + ( β 1 + α 1 )y + γ = 0. The determinant is α 1 + β 1 α + β α + β β 1 + α 1 = α 1 β1 β + α = a b. The advantage of the complex method is obvious. 3. Describe geometrically the sets of points z in the complex plane defined by the following relations. (a) z z 1 = z z, where z 1, z are fixed points in C. This is the midpoint-perpendicular to the segment form z 1 to z. In you do not like this characterization of the points equidistant from z 1 and z, notice first that the midpoint (z 1 + z )/ belongs to the locus we are investigating: z 1 + z z 1 = z z 1 = z 1 + z z. Moreover, setting z 1 = α 1 + iα, z = β 1 + iβ we get for z = x + iy z z 1 = z z (x α 1 ) + (y α ) = (x β 1 ) + (y β ) α 1 + α α 1 x α y = β 1 + β β 1 x β y (β 1 α 1 )x + (β α )y = 1 (β 1 + β α 1 α ). This is the equation of a line with slope λ = β 1 α 1 β α, while the vector z 1 z = (α 1 β 1, α β ) has slope (α β )/(α 1 β 1 ). Their product is 1. So they represent perpendicular lines. (b) 1/z = z. 1/z = z z z = 1 z = 1 z = 1. This is a circle centered at the origin with radius 1. (c) z = R(z) + 1. We set z = x + iy to get z = R(z) + 1 z = (x + 1) x + y = x + x + 1 y = x + 1 = (x + 1/). This is the equation of a parabola with axis the real axis and vertex at ( 1/, 0). 3

4 4. Prove the Lagrange identity n a i b i = i=1 n n a i b i a i bj a j bi. i=1 i=1 1 i<j n We need to show the equivalent formula n a i b i + a i bj a j bi = i=1 1 i<j n n n a i b i. i=1 i=1 We expand out using z = z z to get ( n ) ( n ) a i b i ā j bj + i=1 j=1 = n 1 i<j n (a i bj a j bi )(ā i b j ā j b i ) ( a i b j + a j b i a i ā j b i bj a j ā i b j bi ) a i b i ā j bj + i,j=1 i<j = n a i b i ā j bj + a i b i + ( a i b j + a j b i ) (a i ā j b i bj + a j ā i b j bi ). i j i=j=1 i<j i<j The first and the last sum are equal and, therefore, cancel. This can be seen as follows: In the last sum we have the condition i < j while in the first only i j. A pair (i, j) of unequal integers has either i < j or j < i. In the second case a i b i ā j bj = a j b j ā i bi with j = i < j = i, so we get the term a j b j ā i bi with i < j. The two summands in the middle give exactly n n a i b j i=1 by the distributive law and the same thinking about i j vs i < j and j < i. j=1 5. Prove that a b 1 āb < 1 if a < 1 and b < 1. We have a b 1 āb < 1 a b < 1 āb a b < 1 āb a + b R(a b) < 1+ āb R(āb) (here we notice that R(āb) = R(a b) as they are conjugate numbers) a + b < 1 + a b 0 < 1 + a b a b = (1 a )(1 b ). The assumption a < 1 and b < 1 gives the result. 4

5 6. Prove that it is impossible to define a total ordering on C. In other words, one cannot find a relation between complex numbers so that: (i) For any two complex numbers z and w one and only one of the following is true: z w, w z or z = w. (ii) For all z 1, z, z 3 C the relation z 1 z implies z 1 + z 3 z + z 3. (iii) Moreover, for all z 1, z, z 3 C with z 3 0 Hint: Is i 0? z 1 z = z 1 z 3 z z 3. Since i 0, we have either i 0 or i 0. Step 1: If i 0, then by (iii) we get i i 0 = 0 = 1 0. Since i 0 we also get by (iii) that i 0. We use (ii) to get 0 = i + ( i) 0 + ( i) = i 0. The transitive property gives a contradiction to (i). Step : Assume i 0. Then i 0, since otherwise i 0 and (ii) implies 0 = i + ( i) 0 contradicting (i). As in Step 1, i 0 implies 1 0 and then i 0, which is a contradiction. 7. Prove that the points z 1, z, z 3 are vertices of an equilateral triangle if z 1 +z +z 3 = 0 and z 1 = z = z 3. Rotating the complex numbers z 1, z, z 3 by the same angle keeps their lengths equal and keeps the relation z 1 + z + z 3 = 0 valid. So we can assume that z 1 = r is real. Then r = z = z 3. Setting z = r(cos θ + i sin θ), z 3 = r(cos φ + i sin φ) in polar coordinates we get z 1 + z + z 3 = 0 = r(1 + cos θ + cos φ + i(sin θ + sin φ)) = 0. From the imaginary parts we get sin θ = sin φ = sin θ = sin φ = cos θ = cos φ, by the basic trigonometric identity. This implies that the cosines are either equal or opposite. They cannot be opposite, since the real part of the equation gives 1 + cos θ + cos φ = 0. So 1 + cos θ = 0 = cos θ = 1/ = θ = π/3, 4π/3. 5

6 The equation sin θ = sin φ now gives respectively φ = 4π/3, π/3 (notice that z 3 should be in the left-hand plane to have the same cos as z ). The points are vertices of an equilateral triangle. If you want, you can check that 1 (cos π/3 + i sin π/3) = 1 (cos 4π/3 + i sin 4π/3) = cos π/3 + i sin π/3 (cos 4π/3 + i sin 4π/3). 8. Verify the Cauchy-Riemann equations for z and z 3. We have z = (x y ) + ixy, z 3 = x 3 3xy + i(3x y y 3 ) The first gives for f(z) = z, u = x y, v = xy and = x =, = y =. For the function f(z) = z 3 we have u = x 3 3xy and v = 3x y y 3 and = 3x 3y =, = 6xy =. 9. Express cos(3φ) and cos(4φ) in terms of cos(φ). Express sin(3φ) in terms of sin(φ). By de Moivre s formula (cos φ + i sin φ) n = cos(nφ) + i sin(nφ) for n = 3, 4 we get using the previous exercise: cos(3φ) = R((cos φ + i sin φ) 3 ) = (cos φ) 3 3 cos φ sin φ = cos 3 φ 3 cos φ(1 cos φ) = 4 cos 3 φ 3 cos φ, sin(3φ) = I((cos φ + i sin φ) 3 ) = 3 cos φ sin φ sin 3 φ = 3(1 sin φ) sin φ sin 3 φ = 3 sin φ 4 sin 3 φ. For cos(4φ) we have cos(4φ) = R((cos φ+i sin φ) 4 ) = R(cos φ sin φ+i sin φ cos φ) = (cos φ sin φ) ( sin φ cos φ) = ( cos φ 1) 4 sin φ cos φ = 4 cos 4 φ 4 cos φ+1 4(1 cos φ) cos φ = 8 cos 4 φ 8 cos φ

7 10. Simplify 1 + cos(φ) + cos(φ) + + cos(nφ) and sin(φ) + sin(φ) + + sin(nφ). Set z = cos φ + i sin φ, so that z j = cos(jφ) + i sin(jφ). Then (1 + cos(φ) + cos(φ) + + cos(nφ)) + i(sin(φ) + sin(φ) + + sin(nφ)) = n j=0 (cos(jφ) + i sin(jφ)) = z 0 + z z n = 1 zn+1 1 z We substitute z n+1 = cos(n + 1)φ + i sin(n + 1)φ is the last and multiply with 1 z to get = (1 zn+1 )(1 z) 1 z = We take real and imaginary parts to get (1 cos(n + 1)φ i sin(n + 1)φ)(1 cos φ + i sin φ) (1 cos φ) + sin φ 1+cos(φ)+cos(φ)+ +cos(nφ) = (1 cos(n + 1)φ)(1 cos φ) + sin(n + 1)φ sin φ cos φ = 1 cos φ cos((n + 1)φ)(1 cos φ) + sin((n + 1)φ sin φ) + cos φ cos φ = 1 + cos((n + 1)φ) sin φ/ + sin((n + 1)φ) sin(φ/) cos(φ/) 4 sin φ/ = 1 + cos((n + 1)φ) sin φ/ + sin((n + 1)φ) cos φ/ sin φ/ = 1 + sin((n + 1 φ φ/) sin φ/ = 1 + sin((n + 1/)φ) sin φ/ using the identities sin (x/) = (1 cos x)/, sin x = sin(x/) cos(x/) and the addition theorem for sin. The calculation can be simplified with the use of e iθ. sin φ + sin(φ) + + sin(nφ) = I(1 zn+1 )(1 z) 1 z = = (1 cos((n + 1)φ)) sin φ sin((n + 1)φ)(1 cos φ) cos φ sin φ cos((n + 1)φ) sin φ sin((n + 1)φ)(1 cos φ) 4 sin φ/ = sin(φ/) cos(φ/) cos((n + 1)φ) sin(φ/) cos(φ/) sin((n + 1)φ) sin φ/ 4 sin φ/ = 1 cot φ cos((n + 1)φ) cos φ/ + sin((n + 1)φ) sin φ/ sin φ/ 7 = 1 cot φ cos(n + 1/)φ. sin φ/

8 11. Prove that the diagonals of a parallelogram bisect each other and that the diagonals of a rhombus are orthogonal. If the two sides of the parallelogram are represented by the geometric image of the complex numbers z 1 and z, then the diagonal they enclose is represented by z 1 + z. The midpoint of it is (z 1 + z )/. The second diagonal corresponds to the difference z 1 z. The midpoint of it corresponds to (z 1 z )/. However, the second diagonal is not a vector starting at the origin but at z. So the midpoint of the second diagonal is given by the geometric image of the complex number (z 1 z )/ + z = (z 1 + z )/. The two answers are equal, so the two diagonals meet at their midpoint. We need to show that if z 1 = z then the vectors represented by z 1 + z and z 1 z are perpendicular, or, equivalently, that the arguments of these two complex numbers differ by π/. The difference of the arguments shows up in the quotient, so we need to prove that the complex numbers z 1 + z and z 1 z representing the two diagonals have purely imaginary quotient. But this is true iff z 1 + z z 1 z = z 1 + z z 1 z ( z 1 + z )(z 1 z ) = (z 1 +z )( z 1 z ) z 1 z z 1 z + z z 1 = ( z 1 z z 1 z + z z 1 ) z 1 z = z 1 + z z 1 = z. 1. Prove rigorously that the functions f(z) and f( z) are simultaneously holomorphic. If f(z) = u(x, y)+iv(x, y), then f( z) = u(x, y) iv(x, y). We calculate the partial derivatives with respect to x and y for f( z) using the chain rule from multivariable calculus: (x, y) (x, y) ( v(x, y)) ( v(x, y)) This gives: and = ( v) = (x, y) = (x, y) = ( v) (x, y) (x, y) + + (x, y) ( y) (x, y) ( y) (x, y) + ( v) (x, y) ( y) (x, y) + ( v) (x, y) ( y) = (x, y) (x, y) = (x, y) (x, y) = = = (x, y). = (x, y). = ( v) ( v(x, y)) ( v(x, y)). This proves the equivalence of the Cauchy-Riemann equations. = (x, y). (x, y) = (x, y). 8

9 13. Suppose that U and V are open sets in the complex plane. Prove that if f : U V and g : V C are two functions that are differentiable in the real sense (in x and y) and h = g f, then the complex version of the chain rule is h z = g f z z + g f z z, h z = g f z z + g f z z. Set f(x, y) = u(x, y) + iv(x, y), i.e., (x, y) (u(x, y), v(x, y)) h(u, v). By the standard chain rule for functions of two variables we have h = g + g h = g + g. (1) Moreover, f z f z = 1 ( ) f i f = 1 (u x + iv x i(u y + iv y )) () = 1 ( ) f f i = 1 (u x iv x i(u y iv y )). By the definition of / z and / z we have g = 1 ( ) g z i g g = 1 ( ) g z + i g. We add and subtract the last two equations to get g = g z + g z, g = 1 i ( g z g z We substitute the last equations to (1), multiply the second equation in (1) by i, subtract them to get h = g ( 1 z 1 ) i + g ( 1 1 ) i (3) ). = (g z + g z ) 1 (u x iu y ) + 1 i (g z g z ) 1 (v x iv y ) 1 = g z (u 1 x iu y + iv x + v y ) + g z (u x iu y iv x v y ) = g z f z + g z fz 9

10 using equations (). Similarly we get h = g ( 1 z + 1 ) i + g ( ) i (4) = (g z + g z ) 1 (u x + iu y ) + 1 i (g z g z ) 1 (v x + iv y ) 1 = g z (u 1 x + iu y + iv x v y ) + g z (u x + iu y iv x + v y ) = g z f z + g z f z, since f z f z = 1 ( ) f + i f = 1 (u x + iv x + i(u y + iv y )) = 1 ( ) f f + i = 1 (u x iv x + i(u y iv y )). 14. Show that in polar coordinates the Cauchy-Riemann equations take the form r = 1 r θ, 1 r θ = r. Use these equations to show that the logarithm function defined by log z = log r + iθ, z = r(cos θ + i sin θ), π < θ < π is holomorphic in the region r > 0 and π < θ < π. In polar coordinates x = r cos θ and y = r sin θ. By the chain rule for functions in two variables we have r θ r θ Assume the Cauchy-Riemann equations = cos θ + sin θ (5) = ( r) sin θ + r cos θ = cos θ + sin θ = ( r) sin θ + r cos θ. =, =. (6) 10

11 Then we get r = cos θ sin θ = 1 r θ, (7) θ = ( r sin θ) r cos θ = r r. Conversely assume the Cauchy-Riemann equations in polar form (7). By (5) we get ( r) sin θ + cos θ + sin θ = sin θ + cos θ r cos θ = r cos θ + ( r) sin θ This is a system of linear equations in the unknown functions /, /. Cramer s rule for solving the system gives v x sin θ + v y cos θ sin θ = rv x cos θ rv y sin θ r cos θ cos θ sin θ = rv y r = r sin θ r cos θ = cos θ v x sin θ + v y cos θ r sin θ rv x cos θ rv y sin θ cos θ sin θ r sin θ r cos θ So we get back the Cauchy-Riemann equations. = rv x r =. For the logarithm functions we have u = log r and v = θ. This gives r = 1 r = 1 r θ, θ = 0 = r. 15. Consider the function defined by f(x + iy) = x y, x, y R. Show that f satisfies the Cauchy-Riemann equations at the origin, yet f is not holomorphic at 0. We have f(x + i0) = x 0 = 0, f(0 + iy) = 0 y = 0, f(0 + i0) = 0. Also, if f = u + iv, u = f and v = 0. These give at the origin = lim f(x + i0) f(0 + i0) x 0 x 11 = lim x x = 0,

12 = lim f(0 + iy) f(0 + i0) y 0 x = lim y y Obviously / = 0 and / = 0. So the Cauchy-Riemann equations are satisfied. However, the function is not holomophic at 0. If it were, then f (0) = 0. We set x = y > 0 and let z = x + ix 0. = 0. f(x + iy) f(0 + i0) lim z 0 z = lim z 0 x y z = lim x 0 x x + ix = lim x 0 x x + ix = i. 1