Final #1 - Questions and Solutions

Exercise 1 Let f be an analytic function. It is given that: f(i) = 5 − 2i, Ref(0.01 + i) = 4.97, Ref(−0.02 + 1.01i) = 5.1. Estimate f(−0.01 + 1.02i).

Solution: Writing f(z) = f(x + iy) = u(x, y) + iv(x, y), we may express the data in the form: u(0, 1) = 5, v(0, 1) = −2, u(0.01, 1) = 4.97, u(−0.02, 1.01) = 5.1. Since f is analytic, the partial derivatives u_{x}, u_{y}, v_{x} and v_{y} exist and satisfy Cauchy-Riemann’s equations. Now:

u_{x}(0, 1) ≈ \frac{u(0.01, 1) − u(0, 1)}{0.01} = \frac{4.97 − 5}{0.01} = −3.

Also

u(−0.02, 1.01) ≈ u(0, 1) − 0.02u_{x}(0, 1) + 0.01u_{y}(0, 1),

which yields

u_{y}(0, 1) ≈ \frac{5.1 − 5 + 0.02 · (−3)}{0.01} = 4.

Next

u(−0.01, 1.02) ≈ u(0, 1) − 0.01u_{x}(0, 1) + 0.02u_{y}(0, 1) ≈ 5 − 0.01 · (−3) + 0.02 · 4 = 5.11.

By Cauchy-Riemann’s equations we have:

\begin{matrix}v_{x}(0, 1) = −u_{y}(0, 1) ≈ −4, \\ v_{y}(0, 1) = u_{x}(0, 1) ≈ −3. \end{matrix}

Consequently:

v(−0.01, 1.02) ≈ v(0, 1) − 0.01v_{x}(0, 1) + 0.02v_{y}(0, 1) ≈ −2 − 0.01 · (−4) + 0.02 · (−3) = −2.02.

Thus:

f(−0.01 + 1.02i) = u(−0.01, 1.02) + iv(−0.01, 1.02) ≈ 5.11 − 2.02i.

Exercise 2 Find all solutions of the equation i cos z + sin z = e^{z}.

Solution: Write the equation in the form

e^{z} = i(cos z − i sin z).

The following consecutive steps are self-evident:

e^{z} = ie^{−iz} ,
e^{(1+i)z} = e^{iπ/2} ,

(1 + i)z = \frac{π}{2} · i + 2kπi, k = 0, ±1, ±2, …,

z = \frac{π}{2} · \frac{i}{1 + i} + 2kπ · \frac{i}{1 + i}, k = 0, ±1, ±2, …,

z = (k + \frac{1}{4}) (1 + i)π, k = 0, ±1, ±2, ….

Exercise 3 Evaluate the following integrals:

(a) ∫_{C} \frac{z Log(z + 3)}{(z − i)^{n}} dz , where C is the positively oriented circle of radius 2 around 0 and n

is any integer.

(b) ∫_{C} (e^{z} sin z + e^{z} cos z) dz , where C is the arc y = \sqrt{x}, 0 ≤ x ≤ 1.

Solution: (a) The numerator of the integrand is non-analytic only at the points z on the real axis satisfying z ≤ −3. Hence it is analytic on and inside C. For n ≥ 0 the integrand is analytic on and inside C, whence the integral vanishes. For n = 1, Cauchy’s integral formula yields:

∫_{C} \frac{z Log(z + 3)}{z − i} dz = 2πi[z Log(z + 3)]_{z=i} = −2π Log(3 + i) .

For n ≥ 2 we have

∫_{C} \frac{z Log(z + 3)}{(z − i)^{n}} dz = \frac{2πi}{(n − 1)!} [(z Log(z + 3))^{(n−1)}]_{z=i} .

For n = 2 this gives

∫_{C} \frac{z Log(z + 3)}{(z − i)^{2}} dz = 2πi [\frac{z}{z + 3} + Log(z + 3)]_{z=i} = 2πi [\frac{i}{3 + i} + Log(3 + i)] .

For n ≥ 3:

(z Log(z + 3))^{(n−1)} = \binom{n − 1}{0}z^{(0)} (Log(z + 3))^{(n−1)} + \binom{n − 1}{1}z^{(1)} (Log(z + 3))^{(n−2)}

= z · \frac{(−1)^{n−2}(n − 2)!}{(z + 3)^{n−1}} + (n − 1) · 1 · \frac{(−1)^{n−3}(n − 3)!}{(z + 3)^{n−2}} .

Hence for n ≥ 3:

∫_{C} \frac{z Log(z + 3)}{(z − i)^{n}} dz = \frac{(−1)^{n−1} · 2πi(3n − 3 + i)}{(n − 1)(n − 2)(3 + i)^{n−1}} .

(b)The integrand is an entire analytic function, so that the integral depends only on the endpoints of the arc, and not on the arc itself. One checks easily that F (z) = e^{z} sin z forms a primitive function for the integrand. Therefore:

∫_{C} (e^{z} sin z + e^{z} cos z) dz = [e^{z} sin z]_{z=0}^{1+i} = e^{1+i} sin(1 + i) = \frac{i}{2}(e^{2} − e^{2i}) .

Exercise 4 Prove or disprove:

(a) If R is a rational function, then there exists a square C with an arbitrarily long side for

which ∫_{C} R(z) dz = 0.

(b) If f is an entire non-polynomial function, then it admits a “reasonable” polynomial approximation in the following sense: There exist an M > 0 and a polynomial P such that |f(z) − P (z)| ≤ M for every z ∈ ℂ.

Solution:

(a) True. Let R(z) = P (z)/Q(z) for appropriate polynomials P and Q. Then R is analytic in the whole plane, except for the roots of Q, which are finitely many. Any square C not containing any of these roots on or inside it has the required property.

(b) False. For every polynomial P the function f(z) − P (z) is also entire. This function is non-constant, as otherwise f would be polynomial. By Liouville’s theorem, f(z) − P (z) cannot be bounded.