58.1. PERIODIC FUNCTIONS 1845

Now consider the lines l2 and C. If λ′ (z) = 0 for z∈ l2, a contradiction can be obtained.

Pick such a point. If λ′ (z) = 0, then z is a zero of order m ≥ 2 of the function, λ −λ (z) .

Then by Theorem 52.6.3 there exist δ ,ε > 0 such that if w ∈ B(λ (z) ,δ ) , then λ−1 (w)∩

B(z,ε) contains at least m points.

Ω′

−1

Ω

C l2l1

112

z1 B(z,ε)z

λ (z1)

B(λ (z),δ )

λ (z)

In particular, for z1 ∈ Ω∩B(z,ε) sufficiently close to z,λ (z1) ∈ B(λ (z) ,δ ) and sothe function λ −λ (z1) has at least two distinct zeros. These zeros must be in B(z,ε)∩Ω

because λ (z1) has positive imaginary part and the points on l2 are mapped by λ to a realnumber while the points of B(z,ε) \Ω are mapped by λ to the lower half plane thanks tothe relation, λ (z+2) = λ (z) . This contradicts λ one to one on Ω. Therefore, λ

′ ̸= 0 on l2.Consider C. Points on C are of the form 1− 1

τwhere τ ∈ l2. Therefore, using 58.1.33,

λ

(1− 1

τ

)=

λ (τ)−1λ (τ)

.

Taking the derivative of both sides,

λ′(

1− 1τ

)(1τ2

)=

λ′ (τ)

λ (τ)2 ̸= 0.

Since λ is periodic of period 2 it follows λ′ (z) ̸= 0 for all z ∈ ∪∞

k=−∞(Q+2k) .

Lemma 58.1.25 If Im(τ)> 0 then there exists a unimodular(

a bc d

)such that

c+dτ

a+bτ

is contained in the interior of Q. In fact,∣∣ c+dτ

a+bτ

∣∣≥ 1 and

−1/2≤ Re(

c+dτ

a+bτ

)≤ 1/2.