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1 $begingroup$ Define a mapping $$f(z)=cot(z)-frac {1}{z}-sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$$ Show that $f$ is bounded on $mathbb{C}$ I am trying to derive the Euler factorization of $sin(z)=zprod_{n=1}^{infty}(1-frac {z^2}{n^2pi^2})$ I was able to show that $sum_{n=1}^{infty}frac {2z}{z^2-n^2pi^2}$ converges locally uniformly (i.e. compactly converges on ${zin mathbb{C}:znotin npi, nin mathbb{Z}}$ ), and that $f(z)$ has removable singularities at $npi$ for $nin mathbb{Z}$ , which I will use later to derive the Euler factorization. I am not really sure how to bound $f$ though, and I have a suspicion that $f(z)=0$ . Any help would be much appreciated. Thanks in advance! complex-analysis