# Tangent function Inequality in the complex numbers

Let $$z \in \mathbb{C}.$$

Show there is an $$p>0$$ such that $$|z|. I tried $$\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 $$|z| where $$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 $$|\tan z -\tan 0| = |\tan z| < 1$$.
If you want to find a $$p$$ that works, write $$z=x+iy$$ with $$x,\,y\in\Bbb R$$ and $$t:=\tan x,\,T:=\tanh y$$ so$$1-|\tan^2z|=1-\frac{t^2+T^2}{1+t^2T^2}=\frac{(1-t^2)(1-T^2)}{1+t^2T^2}.$$This is positive provided $$|\tan x|<1$$. If $$0\le p\le\frac{\pi}{4}$$, this holds for each $$z$$ of modulus $$.