Kenneth Kuttler

58.1. PERIODIC FUNCTIONS 1843

fairly near 1 on C1 provided b0 is large so that the circle, C1 has very small radius. Thenalong C, λ (z) is real until it hits C2. What about the behavior of λ on C2? For z ∈ C2, itfollows from the definition of C2 that z = 1− 1

τwhere τ is on the line, a+ ib0. Therefore,

by Lemma 58.1.20, 58.1.17, and 58.1.24

λ (z) = λ

(1− 1

λ(−1

)λ(−1

)−1

=λ( 1

)λ( 1

)−1

=1−λ (τ)

(1−λ (τ))−1=

λ (τ)−1λ (τ)

= 1− 1λ (τ)

which is approximately equal to

1− 116eiπ(a+ib0)

= 1− eπb0e−iaπ

16.

These points are essentially on a large half circle in the upper half plane which has radiusapproximately eπb0

16 .Now let w∈Cwith Im(w) ̸= 0. Then for b0 large enough, the motion over the boundary

of the truncated region indicated in the above picture results in λ tracing out a large simpleclosed curve oriented in the counter clockwise direction which includes w on its interior ifIm(w)> 0 but which excludes w if Im(w)< 0.

Theorem 58.1.22 Let Ω be the domain described above. Then λ maps Ω one to oneand onto the upper half plane of C, {z ∈ C such that Im(z)> 0} . Also, the line λ (l1) =(0,1) ,λ (l2) = (−∞,0) , and λ (C) = (1,∞).

Proof: Let Im(w)> 0 and denote by γ the oriented contour described above and illus-trated in the above picture. Then the winding number of λ ◦ γ about w equals 1. Thus

12πi

∫λ◦γ

1z−w

dz = 1.

But, splitting the contour integrals into l2,the top line, l1,C1,C, and C2 and changing vari-ables on each of these, yields

1 =1

2πi

∫γ

λ′ (z)

λ (z)−wdz

and by the theorem on counting zeros, Theorem 52.6.1 on Page 1663, the function, z→λ (z)−w has exactly one zero inside the truncated Ω. However, this shows this functionhas exactly one zero inside Ω because b0 was arbitrary as long as it is sufficiently large.Since w was an arbitrary element of the upper half plane, this verifies the first assertion ofthe theorem. The remaining claims follow from the above description of λ , in particularthe estimate for λ on C2. This proves the theorem.

Note also that the argument in the above proof shows that if Im(w)< 0, then w is not inλ (Ω) . However, if you consider the reflection of Ω about the y axis, then it will follow thatλ maps this set one to one onto the lower half plane. The argument will make significantuse of Theorem 52.6.3 on Page 1665 which is stated here for convenience.

58.1. PERIODIC FUNCTIONS 1843fairly near | on C; provided bg is large so that the circle, C, has very small radius. Thenalong C, A (z) is real until it hits C). What about the behavior of A on Cy? For z € Co, itfollows from the definition of C2 that z = 1 — ; where T is on the line, a+ ibo. Therefore,by Lemma 58.1.20, 58.1.17, and 58.1.24a(t A(trg = a(i-t)= Me = A)~2G)-1-4G)-1_ 1—A(t) to -1- 1(I-A(z))—1 A(z) A(t)which is approximately equal to1 070 e— ian~ 16e!%(a+ibo) =1- 16These points are essentially on a large half circle in the upper half plane which has radiusapproximately oeNow let w € C with Im (w) 40. Then for bo large enough, the motion over the boundaryof the truncated region indicated in the above picture results in A tracing out a large simpleclosed curve oriented in the counter clockwise direction which includes w on its interior ifIm(w) > 0 but which excludes w if Im(w) < 0.Theorem 58.1.22 Let Q be the domain described above. Then 2 maps Q one to oneand onto the upper half plane of C, {z € C such that Im(z) > 0}. Also, the line A (l,) =(0, 1) A (2) = (—2,0) , and r (C) = (1,).Proof: Let Im(w) > 0 and denote by ¥ the oriented contour described above and illus-trated in the above picture. Then the winding number of A © y about w equals 1. Thus1 120i JAcy Z—WBut, splitting the contour integrals into /2,the top line, 1;,C;,C, and C2 and changing vari-ables on each of these, yields1 uN= 77: | AW 4,2ni Jy A(z) —wand by the theorem on counting zeros, Theorem 52.6.1 on Page 1663, the function, z >A (z) —w has exactly one zero inside the truncated Q. However, this shows this functionhas exactly one zero inside Q because bo was arbitrary as long as it is sufficiently large.Since w was an arbitrary element of the upper half plane, this verifies the first assertion ofthe theorem. The remaining claims follow from the above description of 7, in particularthe estimate for A on C2. This proves the theorem.Note also that the argument in the above proof shows that if Im (w) < 0, then w is not inA (Q) . However, if you consider the reflection of Q about the y axis, then it will follow thatA maps this set one to one onto the lower half plane. The argument will make significantuse of Theorem 52.6.3 on Page 1665 which is stated here for convenience.