Final #2 - Questions and Solutions

Exercise 1 Let f be an analytic function on an open connected set S. It is known that all values f assumes belong to the set {x + ie^{x} : x ∈ ℝ}.

(a) Prove that f is constant on S.

(b) Is the connectedness condition essential?

Solution: (a) Writing f(z) = f(x + iy) = u(x, y) + iv(x, y), the condition in the question means that:

v(x, y) = e^{u(x,y)}, z ∈ S .

By Cauchy-Riemann’s equations:

u_{x}(x, y) = v_{y}(x, y) = \frac{∂}{∂y} e^{u(x,y)} = e^{u(x,y)}u_{y}(x, y).

u_{y}(x, y) = −v_{x}(x, y) = \frac{∂}{∂x}e^{u(x,y)} = −e^{u(x,y)}u_{x}(x, y).  (0.1)

Hence

u_{x}(x, y) = e^{u(x,y)}u_{y}(x, y) = −e^{2u(x,y)}u_{x}(x, y),

and therefore

(1 + e^{2u(x,y)}) u_{x}(x, y) = 0.

The first factor on the left hand side is positive, so that the partial derivative u_{x}(x, y) vanishes identically. By (0.1), so does u_{y}(x, y). Consequently, v_{x}(x, y) and v_{y}(x, y) vanish identically as well. Thus

f′(z) = u_{x}(x, y) + iv_{x}(x, y) ≡ 0,

which implies that f is constant.

(b) The connectedness condition is essential. In fact, let A_{1} and A_{2} be any two open circles with non-intersecting closures, and let S = A_{1} ∪ A_{2}. Define a function f : S ⟶ ℂ by:

f(z) = \{ \begin{matrix}i, & z ∈ A_{1}, \\ 1 + ie, & z ∈ A_{2}. \end{matrix}

The function f is constant in some neighborhood of any point of S, and hence it is analytic throughout S. All values f assumes belong to the set {x + ie^{x} : x ∈ ℝ}, yet the function is not constant.

Exercise 2 Find all solutions of the equation Log (e^{2z} + 1 − i) = z.

Solution: Exponentiating both sides of the equation, we obtain:

e^{2z} + 1 − i = e^{z}.

Set w = e^{z}. Then:

w^{2} − w + 1 − i = 0.

A routine calculation gives:

w_{1} = 1 + i, w_{2} = −i.

The first of these possible values of w leads to:

e^{z} = 1 + i.
e^{z} = \sqrt{2}e^{iπ/4}.

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

The second solution for w gives:

e^{z} = −i.

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

In all the above solutions, the number e^{2z} + 1 − i (which is the same as e^{z}) is not on the non-positive real axis, so that taking the Log is allowed. However, z needs to be in the image of the function Log, which means that its imaginary part has to be between −π and π. Hence the only solutions of the equation are:

z_{1} = \frac{1}{2} ln 2 + \frac{πi}{4}, z_{2} = −\frac{πi}{2} .

Exercise 3 (a) Evaluate the integral ∫_{C} \frac{cos z}{z^{3} − 2iz^{2} − z} dz , where C is the positively ori-

ented circle of radius 2 around 1 + 2i.

(b) Find all possible values of the integral ∫_{C} \frac{dz}{(z^{2} + 4)^{n}} where C is a positively oriented

circle of radius 3, along which z^{2} + 4 ≠ 0, and n is a positive integer.

Solution: (a) The denominator of the integrand factors as:

z^{3} − 2iz^{2} − z = z(z − i)^{2} .

Thus the integrand is analytic everywhere except for the points 0 and i. The point 0 is outside the circle C. Therefore:

∫_{C} \frac{cos z}{z^{3} − 2iz^{2} − z} dz = ∫_{C} \frac{cos z/z}{(z − i)^{2}} dz

= 2πi [(\frac{cos z}{z} )′]_{z=i}

= 2πi [\frac{−z sin z − cos z}{z^{2}} ]_{z=i}

= 2πi(i sin i + cos i) = 2πie^{i·i} = \frac{2πi}{e} .

(b) The integrand is analytic everywhere except for the points ±2i. Hence the integral vanishes unless at least one of these two points lies inside C. If only the point 2i is inside C, then:

∫_{C} \frac{dz}{(z^{2} + 4)^{n}} = ∫_{C} \frac{1/(z + 2i)^{n}}{(z − 2i)^{n}} dz

= \frac{2πi}{(n − 1)!} [(\frac{1}{(z + 2i)^{n}} )^{(n−1)}]_{z=2i}

= \frac{2πi}{(n − 1)!} · [\frac{(−n) · (−n − 1) · . . . · (−n − n − 2)}{(z + 2i)^{2n−1}}]_{z=2i}

= \frac{(−1)^{n−1} · 2πi}{(n − 1)!}(2n − 2) · . . . · (n + 1)n\frac{1}{(4i)^{2n−1}}

= \frac{\binom{2n−2}{n−1}π}{2^{4n−3}} .

If only the point −2i is inside C, then we obtain in a similar fashion:

∫_{C} \frac{dz}{(z^{2} + 4)^{n}} = −\frac{\binom{2n−2}{n−1}π}{2^{4n−3}} .

Finally, if C contains both of these points inside it, then the integral is the sum of the two expressions we obtained, so that it again vanishes.

Exercise 4 Prove or disprove:

(a) Given an anlytic function f in an open region S, for every ε > 0 there exists a δ > 0

such that, for every contour C of length less than δ in S, we have |∫_{C} f(z) dz | < ε .

(b) A non-constant entire analytic function assumes every complex value.

Solution:

(a) False. The function f(z) = 1/z is analytic in ℂ − {0}, but for any simple closed contour

around 0 we have ∫_{C} f(z) dz = 2πi .

(b) False. The function f(z) = e^{z} is a non-constant entire analytic function which does not assume the value 0.