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Compute the maximum of $|f(z)|$ when $|z| \leq 1$ and $f(z)=\sin (z)$

So since $f$ is holomorphic on $|z| \leq 1$, we know we'll find the max of $|f(z)|$ on $|z|=1$.

So:

$|f(z)|=|\sin(z)|=|\frac{e^{iz}-e^{-iz}}{2i}|=|\frac{e^{iz}}{2i}-\frac{e^{-iz}}{2i}|\leq |\frac{e^{iz}}{2i}|+|\frac{e^{-iz}}{2i}|=\frac{|e^{iz}|}{2}+\frac{|e^{-iz}|}{2}=\frac{e^{|iz|}}{2}+\frac{e^{|-iz|}}{2}=e$

I'm not so certain about this solution. Is there another way? Is using the triangle inequality safe when looking for a maximum?

Trio
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2 Answers2

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No, it is not safe, unless you prove that there is a z on the unit circle for which the maximum is attained.

Tom J
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A start: $$|\sin e^{it}| = (1/2)|e^{ie^{it}} - e^{-ie^{it}}| = (1/2)|e^{i(\cos t+i\sin t)} - e^{-i(\cos t+i\sin t)}| \le (1/2)(|e^{i(\cos t+i\sin t)}| + |e^{-i(\cos t+i\sin t)}| ) = (1/2)(e^{-\sin t} + e^{\sin t}).$$

Clearly that can never equal $e.$ So there's further work to do here.

zhw.
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