Representation $\cos(\pi z )$

Question

I already know that $$ \frac{\cos(\pi z)}{\sin(\pi z)}:= \frac{1}{z}+\sum_{n\neq 0} ( \frac{1}{z-n}+\frac{1}{n})\ . $$

However, I am wondering if it also true that

$$ \frac{\cos(\pi z)}{\sin(\pi z)}:= \sum_{n} ( \frac{1}{z-n}+\frac{1}{n+i})\ ? $$

Answer 1

You can re-write your second sum in the form $$ \frac{1}{z} + \frac{1}{i} + \sum\limits_{n \ne 0} {\left( {\frac{1}{{z - n}} + \frac{1}{{n + i}}} \right)} . $$ If your guess was true, subtracting the first formula from the second would give $$ 0 = 1 + 2\sum\limits_{n = 1}^\infty {\frac{1}{{n^2 + 1}}} > 0, $$ which is a contradiction.