11.4. THE LOGARITHM 259

Case 1. a0 ̸= 0. Then for z close to z0,h(z)−1 = 1

a0+∑

∞k=1 bk (z− z0)

k since h(z)−1 hasa derivative, so

f (z) = w+1a0

+∞

∑k=1

bk (z− z0)k = g(z) analytic

which shows the result in this case.Case 2. h(z) = ∑

∞k=m ak (z− z0)

k ,m > 1 where am is the first nonzero a j. Then in thiscase,

h(z)(z− z0)

m =∞

∑k=m

ak (z− z0)k−m ,

(z− z0)m

h(z)= (z− z0)

m ( f (z)−w) =1

am+

∞

∑k=1

ck (z− z0)k ,

for some ck. Thus f (z)−w = 1(z−z0)

m

(1

am+∑

∞k=1 ck (z− z0)

k)= g(z)+∑

mk=1

bk(z−z0)

k for

some analytic function g(z).This shows that if z0 is an isolated singularity, then unless something really bizarre

happens, ( f(B̂)

dense in C) the function has a pole at the singularity or is equal to ananalytic function. How bizarre? If the isolated singularity is not a pole, then f−1 (β )∩ B̂is infinite for every β ∈ C with maybe one exception. This is Picard’s theorem and isan interesting known result but to see this proved see a more advanced book on complexanalysis like Conway [12], Page 300.

11.4 The LogarithmFirst of all, we define for z = x+ iy,ez ≡ ex (cos(y)+ isin(y)) . This agrees with ex wheny = 0. Then it is routine to verify that the usual rules of exponents apply. That d

dz (ez) = ez,

let h = h1 + ih2 and using the power series for cos,sin,

ez+h− ez = ez(

eh1 (cos(h2)+ isin(h2))−1)= ez

(eh1 −1

)+ iezeh1h2 +o(h)

= ezh1 + iezeh1h2 +o(h) = ezh+o(h)

Next I want to define a logarithm which is the inverse of this function. You want to haveelog(z) = z = |z|(cosθ + isinθ) where θ is the angle of z. Now log(z) should be a complexnumber and so it will have a real and imaginary part. Thus

eRe(log(z))+i Im(log(z)) = |z|(cosθ + isinθ) (11.3)

where θ is the angle of z. The magnitude of the left side needs to equal the magnitudeof the right side. Hence, eRe(log(z)) = |z| and so it is clear that Re(log(z)) = ln |z|. Notethat we must exclude z = 0 just as in the real case. What about Im(log(z))? Having foundRe(log(z)) , 11.3 is

|z|(cos(Im(logz))+ isin(Im(logz))) = |z|(cosθ + isinθ) (11.4)

which happens if and only ifIm(logz) = θ +2kπ (11.5)

for k an integer. Thus there are many solutions for Im(logz) to the above problem. Abranch of the logarithm is determined by picking one of them. The idea is that there is only