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Solutions to test i of the math-6640: complex variables and integral transforms with applications course. It includes answers to short answer questions on topics such as the cauchy integral formula, branch points, singularities, and contour integrals.
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MATH-6640 Spring 2005 COMPLEX VARIABLES AND INTEGRAL TRANSFORMS WITH APPLICATIONS TEST I
NAME (print)
(a) The Cauchy Integral Formula says that under certain conditions (fill in the blank on the RHS),
f (z 0 ) =
2 πi
C
f (z) z − z 0 dz
(b) True or false: If f (z) is analytic at each point on a closed contour Γ, then
Γ f^ (z)^ dz^ = 0. Prove or give a counterexample. False. A counterexample is f (z) = 1/z. It is analytic at each point of the unit circle C, yet ∮
C
z dz = 2πi
where C is traversed counterclockwise.
(c) Give an example of a function whose Taylor series diverges at at least one point on the circle of convergence. An example is f (z) = 1/(1 − z) whose Taylor series
n=0 z n (^) converges for |z| < 1 and diverges at z = 1.
(d) State Jordan’s lemma in a form that is appropriate for a large semicircular contour centered at z = 0 and in (i) the upper half plane, and (ii) the left half plane. For (i), Jordan’s lemma states as follows. Consider
IR =
C
f (z)eiαz^ dz,
where α is positive and C is the upper half of the circle z = R. If lim f (z) = 0 as z → ∞ with 0 ≤ arg (z) ≤ π, then limR→∞ IR = 0. For (ii), Jordan’s lemma states as follows. Consider
IR =
C
f (z)eαz^ dz,
where α is positive and C is the left half of the circle z = R. If lim f (z) = 0 as z → ∞ with π/ 2 ≤ arg (z) ≤ 3 π/ 2 , then limR→∞ IR = 0.
(a) (6 points) Prove the Cauchy Estimates: If f (z) is analytic on and inside the circle C : |z−z 0 | = R, and |f (z) ≤ M for all z on C, then
|f (n)(z 0 )| ≤ n!M Rn^ , for n = 1, 2 , 3 , · · ·.
From Cauchy’s extended integral formula we have
f (n)(z 0 ) = n! 2 πi
C
f (z) (z − z 0 )n+1^ dz. In view of the given bounds we have
|f (n)(z 0 )| ≤ n! 2 π
C
|f (z)| |(z − z 0 )n+1| |dz|
= n! 2 π
Rn+
C
|dz|
= n! 2 π
n!M Rn^
(b) (7 points) Suppose g(z) is analytic with a zero of order 4 at z 0. What kind of singularity does f (z) = g′(z)/g(z) have at z 0? Compute the residue of f (z) at z 0. We can write g(z) = (z − z 0 )^4 G(z), where G(z) is analytic and G(z 0 ) 6 = 0. Then,
g′(z) g(z) = (z^ −^ z^0 )
(^4) G′(z) + 4(z − z 0 ) (^3) G(z) (z − z 0 )^4 G(z) = G′(z) G(z)
(z − z 0 )
Here we have ignored the removable singularity at z 0 in the first term on the RHS. The RHS then has a simple pole at z 0 with residue 4.
(c) (7 points) Determine the region in the w -plane to which the function w = Log z maps the z -plane. Choose a suitable branch cut and note that one is considering the principal branch of the logarithm. Let the branch cut be (−∞, 0), so that
Log z = ln |z| + iθ, −π < θ ≤ π.
Consider the ray z = reiα, −π < α ≤ π. Its image in the w -plane is
w = ln r + iα, 0 < r < ∞,
i.e., the horizontal line w = x + iα, −∞ < x < ∞. As α sweeps the interval (−π, π), thereby letting the ray z = reiα^ cover the cut z -plane, the line w = x + iα sweeps the strip − π < Im w < π.
Compute the integral (^) ∫
C
sin z z^4 (z − π/2) dz,
where C is the positively oriented circle |z − 2 | = 3. Show all work. How will the answer change if C is the positively oriented circle |z| = π/2 and the integral is interpreted in the principal value sense?
The integrand has a triple pole at z = 0 and a simple pole at z = π/ 2 , both residing within the contour |z − 2 | = 3. To obtain the residue at z = 0 we expand the function in a Laurent series.
sin z z^4 (z − π/2)
π
2 z π
sin z z^4 = −
πz^4
2 z π
4 z^2 π^2
z − z^3 3!
z^5 5!
πz^3
1 +^2 z π +^4 z
2 π^2
1 − z
2 3!
4 5!
πz^3
2 z π
π^2
z^2 + · · ·
π
z^3
πz^2
π^2
z
We read off the residue as
R 1 = Res
sin z z^4 (z − π/2) , z = 0
π
π^2
The residue at z = π/2 is
R 2 = Res
sin z z^4 (z − π/2) , z^ =^ π/^2
= lim z→π/ 2
sin z z^4 =
π^4. According to the residue theorem,
I =
C
sin z z^4 (z − π/2) dz^ = 2πi^ [R^1 +^ R^2 ] = 2πi
π
π^2 −^
π^4
If C is the circle |z| = π/ 2 , then we consider the indented contour C′^ consisting of C indented with C, a semicircle of radius centered at z = π/ 2. Only the singularity z = 0 lies within C′. Therefore, J =
C
sin z z^4 (z − π/2) dz = 2πiR 1 = − 4 i
π^2
Also, J = I′^ + I, where I′^ is the principal-value integral sought and I is the integral around the semicircle C traversed clockwise. The latter is given by
I = − 1 2 2 πiR 2 = − 16 i π^3
Therefore, I′^ = J − I = − 4 i
π^2
+^16 i π^3
