Prove that $\frac{1}{z+i} +\sin(z)=0$ has infinite solutions over $\mathbb{C}$

Question

Prove that $\frac{1}{z+i} +\sin(z)=0$ has infinite solutions over $\mathbb{C}$

Can someone give me a clue?

Answer 1

You can use Rouché's theorem.

Spoiler:

Notice that $\forall_{n \in \Bbb N} \sin(2 \pi n)=0$.

For each $n$ and for $\theta \in [0,2 \pi]$ we have that $\sin(2 \pi n + re^{i \theta}) = \sin(2 \pi n)\cos(re^{i \theta}) + \cos(2 \pi n)\sin(re^{i \theta}) = \sin(re^{i \theta}) = \sin(r(-1)^{\frac{\theta}{\pi}})$. Hence, we can choose $r > 0$ such that $|\sin(2 \pi n + re^{i \theta})| > \epsilon > 0$.

There exists an $N$, such that $\forall_{n > N}|\frac{1}{(2 \pi n + re^{i \theta})+i}| < \epsilon$.

By applying Rouché's theorem for each $n>N$, we find that $\frac{1}{z+i} + \sin(z)$ has infinite zeros.