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Prove that if $\left| \sin z \right| \leq 1$, then $$ \left|y\right| \leq \ln\left(\sqrt{2}+1\right) .$$

I am not familiar with hyperbolic function before learning complex analysis but in order to deal with questions like this, hyperbolic functions do seem useful!

For example,

$$\sin (x+iy) = \sin x \cosh y + i \sinh y \cos x$$

With further simplification, we can also show

$$\left| \sin z \right| = \sqrt{\cosh^2y - \cos ^2 x}$$

I get the sense that I am close, but how should I get the range of $y$ from these? Thanks in advance.

Travis Willse
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Nighty
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2 Answers2

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HINT:

$$\implies\cosh^2y-\cos^2x\le1$$

$$\implies\cosh^2y\le\cos^2x+1\le1+1$$

As $\cosh y\ge1,$

$$\implies\cosh y\le\sqrt2$$

Now $2\cosh y=e^y+e^{-y}$

lab bhattacharjee
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You're almost there: Squaring and rearranging gives $$\cosh^2 y = |\sin z|^2 + \cos^2 x, $$ but $0 \leq \cos^2 x \leq 1$ and by hypothesis $|\sin z|^2 \leq 1$, so $$\cosh^2 y \leq 1 + 1 = 2 ,$$ and hence $$1 \leq \cosh y \leq \sqrt{2}. $$ Now, use that $$\operatorname{arcosh} u = \log\left(u + \sqrt{u^2 - 1}\right) ,$$ which in particular is an increasing function.

Travis Willse
  • 99,363
  • How do we obtain the last equality? – Nighty Oct 25 '15 at 13:55
  • Since $\operatorname{arcosh}$ is increasing (to see this, we can simply show that its derivative is positive), we have that $$\operatorname{arcosh} \cosh y \leq \operatorname{arcosh}(\sqrt{2}) = \log\left[\left(\sqrt{2}\right) + \sqrt{\left(\sqrt{2}\right)^2 - 1} \right] = \log\left(\sqrt{2} + 1\right).$$ Since $\operatorname{arcosh} \cosh y$ is $y$ for $y \geq 0$ and $-y$ for $y \leq 0$, we have $\operatorname{arcosh} \cosh y = |y|$. – Travis Willse Oct 25 '15 at 20:24