Tangent function Inequality in the complex numbers Let $z \in \mathbb{C}.$
Show there is an $p>0$ such that $|z|<p \implies |\tan(z)|<1$.
I tried $\tan(z)=\frac{\sin(z)}{\cos(z)}$ and replacing $\cos(z)$ and $\sin(z)$ by its exponential forms. Then I took the absolute value of $\tan(z)$ but it's not taking me anywhere. I realize that in $\mathbb{R},$ $p=\frac{\pi}{4}$ I think, right? So I guess it's the same for the complex numbers? But how do I prove this?
I have another question related to this which is proving that $|z|<p \implies a(\tan(z))=z$ where $$a(z)=\sum_{n=0}^{\infty} (-1)^n\frac{z^{2n+1}}{(2n+1)}$$
I already proved that $a(z)$ has radius of convergence 1, if that helps...

The tangent function is continuous. In particular, it is continuous at $z=0$. From the definition of continuity taking $\epsilon = 1$, there is some $\delta >0$ such that if $|z-0| < \delta$, then $|\tan z -\tan 0| = |\tan z| < 1$.
