Final answer:
To show that C(z) is bounded within the square γ_N, we rely on the fact that the cotangent function has singularities only at integer multiples of π and the square's vertices are at half-integer points, guaranteeing a fixed distance from the singularities.
Step-by-step explanation:
The question is asking to demonstrate that the function C(z) = πcot πz is bounded on the square γ_N with vertices at ±(N+1/2) ± i(N+1/2). To show this, we first need to understand the behavior of the cotangent function. Cotangent has singularities, points where the function heads to infinity, at every integer multiple of π. However, on the square defined by γ_N, none of these singularities are present, since the vertices are at half-integer points.
Furthermore, the magnitude of the cotangent function outside of singularities and within the region bounded away from them is limited. Since the vertices of the square γ_N are a fixed distance away from the nearest singularities, the function C(z) will attain a maximum value within this square that does not depend on the size of the square, and hence the value of N. Even as N grows infinitely large, the maximum value of C(z) remains the same because the distance to the nearest singularity remains unchanged. Thus, within the square γ_N, C(z) is bounded by a value that does not depend on N.