Patrick Corn
Math 185, 8/9/05
Solution Set 12
Problem 1: Suppose that f(z) = (z−z0)mg(z) where g(z) is holomorphic and nonzero in a
neighborhood of z0. Then, as we have seen a few times already, the residue of f′/f at z0is m.
(Just use the product rule.) So if fhas a zero of order k, the residue at it is k, and if fhas a pole
of order ℓ, the residue at it is −ℓ. Now just apply the residue theorem.
Problem 2: Let z0be a zero of f. Let Crbe the circle of radius raround z0, where ris small
enough so that fhas no zeroes other than z0on or inside Cr. Let ǫ= inf{|f(z)|:z∈Cr}. Then
ǫ > 0, and we can find N∈Nsuch that |f(z)−fn(z)|< ǫ ≤ |f(z)|for all n > N and z∈Cr. So
then fand fnhave the same number of zeroes inside Cr(with multiplicities), by Rouch´e’s theorem.
Doing this process at each of the mzeroes, we see that fnhas at least mzeroes if fhas at least m
zeroes, for ngreater than the maximum of the numbers Nwe obtain via the above process.
As for the corollary, fix w∈G, and consider the functions fn(z)−f(w), which converge to
f(z)−f(w). If the latter function has at least two zeroes, then fn(z)−f(w) has at least two zeroes
for sufficiently large n, but this contradicts the hypothesis that fnis univalent. So f(z)−f(w) has
one zero for all w∈G, so fis univalent.
Problem 3: Let L1be the line segment {it :|t| ≤ R}. Let L2be the arc {Reit :π/2≤t≤3π /2}.
Let Kbe the closed half-disk bounded by L1and L2. Let f(z) = z+a. Let g(z) = z+a−ez.
Suppose Ris large enough so that the closed disk |z+a| ≤ 1 is contained in the interior of K.
Then on L1and L2, we have
|f(z)−g(z)|=|ez| ≤ 1<|z+a|=|f(z)|,
and f(z) has exactly one zero in K, so g(z) has exactly one zero there as well. As we let Rget
bigger, Kincreases to cover the entire left half plane, so the result follows. (If there was another
zero, it would lie inside Kfor large enough R, which would be a contradiction of the above Rouch´e’s
Theorem argument.)
To see that this zero is on the real axis, note that g(−a) = −e−a<0 and g(0) = a−1>0, so g
has a zero in (−a, 0) by the Intermediate Value Theorem.
Problem 4: (i) The LFT φ1(z) = −i(z−1)
z+1 should work.
(ii) The function φ2(z) = −φ(z)2should work, since squaring sends the upper half plane to C
minus the nonnegative real axis.
(iii) Try φ3(z) = Log(φ2(z))+πi
2π.
(iv) Try φ4(z) = pφ1(z) = eLog(φ1(z))/2. You might also use eπφ3(z)/2, if you liked doing things
sequentially.
(v) Try φ5(z) = 1−φ4(z)
1+φ4(z).
(vi) Try φ6(z) = φ5(z)2. If you want an explicit formula for φ6, you can get it if you want; knock
yourself out.
Problem 5: Suppose f(0) = 0, f′(0) is real and positive, and fmaps the open unit disk onto
itself. Well, then g=f−1has the same properties (because g′and f′are reciprocals). In fact, by
Schwarz’s lemma, both f′(0) and g′(0) are ≤1, but they are reciprocals, so they must both equal
1. Since equality holds, f(z) = λz, but λ=f′(0) = 1, so we are done.
Problem 6: Let fbe a conformal automorphism of C, and consider the singularity of fat ∞.
We have seen that if it is removable, then fis a constant, which is no good. We have also seen
that if it is a pole, then fis a nonconstant polynomial. But the univalence of fimplies that it
factors as a(z−b)nfor some n(because it can’t have more than one zero, but it has to factor
1
