58.1. PERIODIC FUNCTIONS 1823

Consider the claim about the existence of an inverse. Let ad− cb ̸= 0 for f (z) = az+bcz+d .

Then

f (z) = φ

((a bc d

))It follows

(a bc d

)−1

exists and equals 1ad−bc

(d −b−c a

). Therefore,

z = φ (I)(z) = φ

((a bc d

)(1

ad−bc

(d −b−c a

)))(z)

= φ

((a bc d

))◦φ

((1

ad−bc

(d −b−c a

)))(z)

= f ◦ f−1 (z)

which shows f−1 exists and it is easy to find.Next suppose f−1 exists. I need to verify the condition ad− cb ̸= 0. If f−1 exists, then

from the process used to find it, you see that it must be a fractional linear transformation.

Letting A =

(a bc d

)so φ (A) = f , it follows there exists a matrix B such that

φ (BA)(z) = φ (B)◦φ (A)(z) = z.

However, it was shown that this implies BA is a nonzero multiple of I which requires thatA−1 must exist. Hence the condition must hold.

Theorem 58.1.12 If f is a nonconstant elliptic function with a basis {w1,w2} for the mod-ule of periods, then {w′1,w′2} is another basis, if and only if there exists a unimodular

transformation,(

a bc d

)= A such that(

w′1w′2

)=

(a bc d

)(w1w2

). (58.1.4)

Proof: Since {w1,w2} is a basis, there exist integers, a,b,c,d such that 58.1.4 holds.It remains to show the transformation determined by the matrix is unimodular. Takingconjugates, (

w′1w′2

)=

(a bc d

)(w1w2

).

Therefore, (w′1 w′1w′2 w′2

)=

(a bc d

)(w1 w1w2 w2

)Now since {w′1,w′2} is also given to be a basis, there exits another matrix having all integer

entries,(

e fg h

)such that

(w1w2

)=

(e fg h

)(w′1w′2

)