MAT334 Complex Variables Homework 4 Solutions
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MAT334 Complex Variables
Homework 4 Solutions
2022
1. Let γ 1 be the line segment from i to 2 + i and γ2 be the semicircle from i to 2 + i passing through 2i + 1. (1a) Parameterize γ 1 and γ2 .
Notice that γ2 is half of the circle centered at i + 1 with radius 1. We parameterize:
γ 1 (t) =i + t t ∈ [0, 2]
γ2 (t) = 1 + i − e−it t ∈ [0, π]
(1b) Using your parameterization in part (a), evaluate lγ1 z( +z¯ 1)dz and lγ2 z(z¯ + 1)dz .
By definition,
l z( +z¯ 1)dz = l 2 (i + t)(−i + t + 1)dt
= l02 t2 + t + i + 1dt
8
3
And,
lγ2 z( +z¯ 1)dz = l0π (1 + i − e−it)(2 − i − eit )(ie−it)dt
= l0π (1 − i) − (1 − 4i)e−it − (1 + 2i)e−2it dt
= l0π (1 − i) − (1 − 4i)(cos(−t) + i sin(−t))− (1 + 2i)(cos(−2t) + i sin(−2t))dt
= l0π ( 1 + cos(t) + 4 sin(t) − cos(2t) − 2 sin(2t))+ i (− 1 − 4 cos(t) − sin(t) − 2 cos(2t) − sin(2t))dt
(1c) Let γ3 be γ1 − γ2 and evaluate lγ3 z( +z¯ 1)dz and lγ3 dz .
• We have that 1γ3 z(z¯+ 1)dz = 1γ1 z(z¯+ 1)dz− 1γ2 z(z¯+ 1)dz = + 4 + 2i− (8 +π +(2 − π)i) = − − π +πi .
• Furthermore, note that γ3 is a piecewise smooth, positively oriented, simple, closed curve.
• Therefore, 1i(+)i) dz = 0 by the Cauchy Integral Theorem.
2. Let γ : [0, 1] → C be the curve given by γ(t) = i(1 − t) − 2it . Show that 'lγ cos(z)dz ' < 3e2 (Hint: Use question 4 on HW2).
• Note that γ(t) = i − 3it .
• Let L = {z ∈ C : Re z = 0}. We know from question 4 on HW2 that for all z ∈ L we have | cos(z)| ≤ e| Imz| . Further, note that γ(t) ∈ L for all t ∈ [0, 1].
• For t ∈[0, ], we have | Im(γ(t))| = |γ(t)| ≤ 1. For t ∈[, 1], we have | Im(γ(t))| = |γ(t)| ≤ 2.
• Therefore, for t ∈[0, ], | cos(z)| ≤ e and for t ∈[, 1], | cos(z)| ≤ e2 .
e([0),ine(32])s([),1 ] . Note that γ 1 has length 2 and γ2 has length 1. 'lγ cos(z)dz ≤' 'lγ1 cos(z)dz +' 'lγ2 cos(z)dz '
• Since | cos(z)| ≤ e on γ1 and | cos(z)| ≤ e2 on γ2 , we have that M1 = max {| cos(γ1 (t))| : t ∈[0, ]} ≤ e and M2 = max {| cos(γw (t))| : t ∈[, 1]} ≤ e2 .
• ML estimation now gives us
'lγ cos(z)dz ≤' 'lγ1 cos(z)dz +' 'lγ2 cos(z)dz ' ≤ 2e + e2 < 3e2
3. For a curve γ : [a, b] → C, define γR : [a, b] → R2 by γR(t) = (Re(γ(t)), Im(γ(t)). For a function f : C → C, write f = u + iv and define functions g , h : R2 → R2 by g(x , y) = (u(x , y), −v(x , y)) and h(x , y) = (v(x , y), u(x , y)).
(3a) Write lγR g · dγ and lγR h · dγ in terms of lγ f dz .
• By the definition of the line integral for vector fields, lγR g · dγ = lγR udx − vdy and lγR g · dγ = lγR vdx + udy .
• By the definition of the line integral for complex functions, lγ f dz = lγR udx − vdy +lγR vdx + udy .
• Therefore, lγR g · dγ = Re (lγ f dz)and lγR h · dγ = Im (lγ f dz).
(3b) Prove that if f is entire, then g is conservative.
• Since f is entire, uy = −vx .
• Since g = (u, −v), we have that ∂1 g2 = −vx = uy ∂2 g1 on R2 .
• From your second year calculus class, we know that since ∂1 g2 = ∂2 g1 on R2 , then g is conservative on R2 .
(3c) If g is conservative, is f entire? Prove your answer.
• No. Consider the following counterexample.
• Let f (z) = z . Note for x + iy ∈ C, f (x + iy) = x − iy so g(x , y) = (x , y).
• By Cauchy-Riemann, f is not differentiable anywhere, and is not entire.
• However, g is conservative. There are several ways to see this. We could note that ∂1 g2 = 0 = ∂2 g1 or that g has a potential function p(x , y) = .
4. Let U ⊂ C be open and f : U → C be analytic. For n ∈ N, we define an nth order primitive for f on U to be any function F : U → C such that = f .
Prove that if f is entire, then f has an nth order primitive for all n ∈ N.
• We proceed by induction.
• Since f is entire, it is analytic on the simply-connected domain C. By theorem 3.2.8 in the course notes, f has a first order primitive on C.
• Assume that f has a kth order primitive F on C for some k ∈ N.
• Since is defined on C and k ≥ 1, we have that is defined on C. Therefore, F is entire.
• By the n = 1 case, F has a first order primitive on C.
• Then = = f . So is a k + 1st order primitive for f on C.
• Therefore, f has an nth order primitive on C for all n ∈ N by induction.
2022-08-02