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Show that if $f$ is analytic on disc $D(0,1)$ and $f(x)=\tan x$ , when $0 < x < 1$

then there is no $z$ in $D$ with $f(z)=i$.

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    yes.. the open disc with center$0$ and radius 12012-11-09

1 Answers 1

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Using the identity theorem we get $f(z)=\tan z$ for all $z \in D$. If $\tan z=i$ for some $z$ we have $\sin z=i \cos z$. Now we use $\sin^2 z+ \cos^2 z=1$ to get $1=0$. A contradiction.