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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...

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2 Answers 2

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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 $|\tan z -\tan 0| = |\tan z| < 1$.

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  • $\begingroup$ Thank you! Didn't think of that. $\endgroup$ Aug 13, 2020 at 18:06
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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 $<p$.

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