Math 503 Midterm Exam I Solution Set

October 2000 / Ali Sinan Sertöz

Question 1: Let f : (X, d) ⟶ (Ω, ρ) be a continuous function from the metric space X into the metric space Ω. Assume that every open cover of X admits a finite subcover. Show that f is uniformly continuous.

Solution 1: (This is an exercise I mentioned in class. In particular I suggested that may be the δ’s and ϵ’s involved should be divided by 2.) f is continuous at every x ∈ X, so ∀ϵ > 0 ∃δ_{x} > 0 such that d(x, y) < δ_{x} ⟹ ρ(f(x), f(y)) < ϵ/2. Let U = {B(x, δ_{x}/2)|x ∈ X} be an open cover of X. Let U′ = {B(x_{1}, δ_{x_{1}}/2), …, B(x_{n}, δ_{x_{n}}/2)} be a finite subcover. Let 0 < δ ≤ min{δ_{x_{1}}/2, …, δ_{x_{n}}/2}. Now for any p, q ∈ X with d(p, q) < δ we want to show that ρ(f(p), f(q)) < ϵ. For this observe that since U′ is an open cover for X, there is an index i such that p ∈ B(x_{i}, δ_{x_{i}}/2), i.e. d(p, x_{i}) < δ_{x_{i}}/2. From the continuity of f at x_{i} we have ρ(f(p), f(x_{i})) < ϵ/2. Also observe that d(q, x_{i}) ≤ d(q, p) + d(p, x_{i}) < δ_{x_{i}}/2 + δ_{x_{i}}/2 = δ_{x_{i}}. Again from the continuity of f at x_{i} we have ρ(f(q), f(x_{i})) < ϵ/2. Putting these together; ρ(f(p), f(q)) ≤ ρ(f(p), f(x_{i})) + ρ(f(x_{i}), f(q)) < ϵ/2 + ϵ/2 = ϵ, thus we achieve what we wanted to show.

Question 2: Let z_{1}, …, z_{n} be complex numbers such that Re(z_{k}) > 0 and Re(z_{1} ⋯ z_{k}) > 0 for all k = 1, …, n. Show that log(z_{1} ⋯ z_{n}) = log(z_{1}) + ⋯ + log(z_{n}), where log denotes the principal branch. If the restrictions on the z_{k} are removed, does the formula remain valid?

Solution 2: (This is an exercise from the book; page 44, Exercise 20.) Let z_{k} = r_{k}e^{iθ_{k}} where r_{k} ∈ ℝ^{+} and −π < θ_{k} < π, k = 1, …, n. The other restrictions on the z_{k}’s imply that −π < θ_{1} + ⋯ + θ_{k} < π for all k = 1, …, n. To see this, note that z_{1} ⋯ z_{k} = (r_{1} ⋯ r_{k})e^{i(θ_{1}+···+θ_{k})} = (r_{1} ⋯ r_{k}) cos(θ_{1} + ⋯ + θ_{k}) + i(r_{1} ⋯ r_{k}) sin(θ_{1} + ⋯ + θ_{k}). Hence Re(z_{1} ⋯ z_{k}) = (r_{1} ⋯ r_{k}) cos(θ_{1} + ⋯ + θ_{k}) > 0 implies −π < θ_{1} + ⋯ + θ_{k} < π. This shows that the arguments of all the numbers involved are in the domain of the principal branch of the log function. Hence log(z_{1} ⋯ z_{n}) = log ((r_{1} ⋯ r_{n})e^{i(θ_{1}+···+θ_{n})}) = log(r_{1} ⋯ r_{n}) + i(θ_{1} + ⋯ + θ_{n}) = log r_{1} + ⋯ + log r_{n} + iθ_{1} + ⋯ + iθ_{n} = log z_{1} + ⋯ + log z_{n} Clearly if the restrictions on the z_{k} are removed the sum of the arguments may exceed π and the claimed equality no longer holds. For example let z_{1} = z_{2} = z_{3} = z_{4} = i Then log(z_{1}z_{2}z_{3}z_{4}) = log 1 = 0 but log(z_{1}) + ⋯ + log(z_{4}) = iπ/2 + ⋯ + iπ/2 = 2πi ≠ 0.

Question 3: Let T be a Möbius transformation of the form \frac{az + b}{cz + d} where c ≠ 0. Find all Möbius

transformations which commute with T.

Solution 3: (This was intended to be exercise 24 on page 57 of the book but in my efforts to hide its source I managed to increase the scope of the problem slightly above what I had in mind. But anyone who has at least read the exercise would know where to start. The condition c ≠ 0 is intended to leave out infinity from the fixed point set.) First let us show that if T and S have the same fixed points then they commute. Let T and S have the fixed points α, β ∈ ℂ. First assume that α ≠ β. Any Möbius

transformation T having these fixed points satisfies the equation \frac{T (z) − α}{T (z) − β} = k_{T} \frac{z − α}{z − β} for some k_{T} ∈ ℂ.

Solving for T (z) we find T (z) = \frac{(α − βk_{T})z + αβ(k_{T} − 1)}{(1 − k_{T})z + αk_{T} − β}. Hence T has the matrix representation

T = (\begin{matrix}(α − βk_{T}) & αβ(k_{T} − 1) \\ (1 − k_{T}) & αk_{T} − β \end{matrix}). Similarly for S with the constant k_{S}. It is now straightforward to

check, using the matrix representations, that S and T commute. Next assume that α = β. Then any

such Möbius transformation satisfies the equation \frac{k_{T}}{T (z) − α} = \frac{k_{T}}{z − α} + u_{T} for some constants k_{T} , u_{T} ∈

ℂ. Solving for T (z) gives T (z) = \frac{(k_{T} + αu_{T})z − α^{2}u_{T}}{u_{T} z + k_{T} − αu_{T}} . Hence T has the matrix representation T =

(\begin{matrix}(k_{T} + αu_{T}) & −α^{2}u_{T} \\ u_{T} & k_{T} − αu_{T} \end{matrix} ). Similarly for S with the constants k_{S}, u_{S}. Again it is straightforward to

check as above that S and T commute. So far is the scope of the exercise from the book. i.e. we showed that if two transformations have the same fixed points then they commute. To thoroughly examine the commutation behaviour requires a little more calculation. Here we quote the following result: Let T have two distinct fixed points and T^{2} not the identity map. Then S commutes with T if and only if S has the same fixed points. If T has only one fixed point then S commutes with T if and only if S has only one fixed point which is the fixed point of T. And finally if T^{2} is the identity map then S commutes with T if and only if either S has the same fixed point(s) as T or S interchanges the fixed points of T.

Question 4: Find the principal branch and the range of ω = (3 − z)^{\frac{1}{3}} .

Solution 4: (This is solved in exactly the same way as the example I did in class, which was the same question with ω = (1 − z)^{\frac{1}{2}} .) First note that ω = exp(\frac{1}{3} log(3 − z)). Since the principal branch of log excludes the negative real axis together with the origin, we should exclude all real z ≥ 3. Then log takes 3 − z onto the strip between iπ and −iπ. Then \frac{1}{3} narrows the strip down to between \frac{iπ}{3} and \frac{−iπ}{3} . Finally exponentiation takes this to the wedge around the real axis −\frac{π}{3} < arg ω < \frac{π}{3} .

Question 5: Calculate the principal values of 2^{i}, i^{i} and i^{i^{i}}. Also calculate the cross-ratio (1, i, −1, −2i). Write your answers in the simplified form a + ib with a, b ∈ ℝ.

Solution 5: (You cannot claim you are taking a course in complex analysis if you cannot do arithmetic with complex numbers. Such calculations are second nature to every mathematician.)

2^{i} = exp(i log 2) = cos(log 2) + i sin(log 2)

i^{i} = exp(i log i) = exp(i log e^{i \frac{π}{2}} ) = exp(i · i\frac{π}{2} ) = e^{− \frac{π}{2}} .

i^{i^{i}} = exp(i^{i} log i) = exp(e^{− \frac{π}{2}} · i\frac{π}{2} ) = cos(\frac{π}{2} e^{− \frac{π}{2}} ) + i sin(\frac{π}{2} e^{− \frac{π}{2}} ).

(1, i, −1, −2i) = \frac{1+1}{1+2i} \frac{i+2i}{i+1} = \frac{6i}{−1+3i} = \frac{6i}{−1+3i} · \frac{−1−3i}{−1−3i} = \frac{18−6i}{10} = \frac{9}{5} − i\frac{3}{5}.