Math 407 Solutions to Exam 3 April 24, Let f = u + iv, and let Γ be parmetrized by Γ = (x (t),y(t)) = x (t) +iy (t) for a t b.

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1 Math 7 Solutions to Exam 3 April, 3 1. (1 Define the following: (a f (z dz, where Γ is a smooth curve with finite length. Γ Let f u + iv, and let Γ be parmetrized by Γ (x (t,y(t x (t +iy (t for a t b. Then Γ f (z dz b a ( u dx dt v dy b ( dt + i u dy dt dt + v dx dt dt a The parametrization of Γ is assumed to be smooth and the functions u and v are assumed to be continuous functions of x and y. (b The function f is analytic at z. f is analytic at z if there is a neighborhood of z such that f is differentiable at every point in this neighborhood. (c The function f has a pole of order at z. The statement means that z is an isolated singular point of f, and that the Laurent expansion of f about the point z has the form f (z b (z z + b 3 (z z 3 b (z z + b 1 (z z 1 + a n (z z n n with the coefficienct b. (d The residue of f at z. The residue of f at z is the coefficient of (z z 1 in the Laurent expansion of f.. (8 If f (z u+iv has a derivative at z x + iy, then the real valued functions u and v satisfy a pair of partial differential equations at z. These equations are called the Cauchy-Riemann equations. (a State the Cauchy-Riemann equations. x v y v y x

2 (b Prove that u and v must satisfy these equations. The Cauchy Riemann equations follow from the fact that if f (z exists, then its value must be the same regardless of how z appoaches zero. The lines below calculate the derivative of f by first letting z appoaches zero through real values and then through imaginary values. Set f u + iv, and z h, with h a real number, then df dz h u (x + h, y+iv (x + h, y u (x, y iv (x, y h u (x + h, y u (x, y iv (x + h, y iv (x, y + lim h h h h x + i v x Now set z ih, with h a real number, then df dz u (x, y + h+iv (x, y + h u (x, y iv (x, y h ih u (x, y + h u (x, y iv (x, y + h iv (x, y + lim h ih h ih 1 i y + v y v y i y Two complex numbers are equal if and only if they have the same real and imaginary parts. Thus, x v y v y x

3 3. (15 (a State the Cauchy-Goursat theorem. Be sure to include any needed hypotheses. Suppose f (z is analytic inside and on a simple closed contour Γ. Then f (z dz Γ (b Suppose f has singular points z 1 and z inside a closed contour C. Why does C f (z dz πi (Res zz1 f + Res zz f Let C 1 and C denote positively oriented circles centered at z 1 and z respectively, such that both circles and their interiors are disjoint from each other, and both are completely contained inside the original contour C. Then the Cauchy-Goursat theorem is used to show that f (z dz f (z dz + f (z dz C C 1 C The value of the contour integral of f over these positively oriented circles is then computed by looking at the Laurent series expansion of f around each of these points. We parametrize the circle C i by z (t z i +ρe iθ, θ π. Then we have ( f (z dz c n (z z i n dz C i C i n n n c n C i (z z i n dz π c n i ρ n+1 e i(n+1θ dθ The only one of the integrals that is not equal to zero is the one in which the exponent in e i(n+1θ is zero. That is, n +1 orn 1. Thus, π f (z dz (c 1 i dθ C i πi (c 1 πires zzi f 3

4 . ( Let f (z z (z 1 (z. (a Locate all of the isolated singular points of f and determine their type. The isolated singular points of f are z, 1, and. Since is zero at z, and not equal to zero at the other two singular points, is a removable singularity, z 1 is a simple pole, that is a pole of order 1, and z is a pole of order. (b Calculate the residue of f at each of its isolated singular points. At z the residue of f is zero. At z 1 the residue equals At z the residue equals Res z1 f (z 1 z 1 z (z 1 (z d Res z f z dz z 1 z (z sin [ (z [ d z dz z (z 1 z (z 1 (z ] z (cos z z (z 1 (z 1 z (z 1 cos 3 sin.89 ]

5 5. ( Let f (z z ( e 1/z 1. (a Find the Laurent series expansion of f about z. What type of singular point is? ( f (z z 1 z n n! 1 n1 n 1 z n n! z z (3! + 1 z (! z n (n+! + Since the number of terms in which z appears to a negative exponent is not finite, is an essential singular point for f. (b Let C denote the positively oriented circle of radius centered at the origin. Calculate C f (z dz. f (z dz πi (Res z f C πi ( 1 6 πi 3 6. (15 Let f (z z 1+z and let C R denote the upper half of the positively oriented circle of radius R centered at the origin. (a Show lim C R f(z dz. If z R then z 1+z R 1+z f (z dz C R R R 1. Thus, R R 1 (πr πr3 R 1 Thus, as R gets arbitrarily large the absolute value of the integral approaches zero. That is, C R f(z dz. lim 5

6 (b Using residue theory calculate 1+xdx. Hint: the integrand is an even function of x. Let Γ R denote the closed contour that consists of that part of the real line between R and R, and the half circle C R. Then we have 1+x dx 1 1 lim 1+x dx ( R R 1 lim z (ΓR 1+z dz 1+x CR dx + z 1+z dz For R>1 the only singular points of the integrand are at z 1 e πi/ and z e 3πi /. These are simple poles of the integrand. Thus, 1+x dx 1 [ [ ] z z (Res πi zz1 ]+ 1+z Res zz 1+z ( z πi 1 z1 3 + z z 3 πi ( z 1 z πi ( e πi/ + e 3πi πi ( ( π i sin π 6

7 7. (1 Let C ρ denote the top half of the circle of radius ρ centered at the origin. Suppose that C ρ is traced out from right to left. Calculate lim ρ + C ρ z dz. The upper half circle is parametrized by z ρe iθ, θ π. The Laurent series expansion of /z about z is Thus, the contour integral equals C ρ z dz z 1 z n n ( 1 n z n+1 (n +1! ( 1 n z n 1 (n +1! 1 z + ( 1 n z n 1 (n +1! π π πi [ n1 1 ρe iθ + idθ+i n 1 n1 ( ρe ( 1 n iθ ] n 1 iρe iθ dθ (n +1! ( 1 n ρ n (n + 1! π e niθ dθ Note: for this particular function there is no need to actually take the limit as ρ tends to zero from above as each of the integrals π eniθ dθ. Thus, lim ρ + C ρ z dz. πi 7