sin (z) = a and a\in [-1,1]

**dwsmith** · December 22nd 2010, 07:08 PM

If $\sin{z}=a, \ a\in [-1,1]$ , then what can be said about $z\mbox{?}$

Since a is between -1 and 1, $a=\cos{\theta}$ .

$\sin{z}=\sin{x}\cosh{y}+\mathbf{i}\cos{x}\sinh{y}$

Setting the real and imaginary parts equal we obtain:

$\displaystyle\sin{x}\cosh{y}=\cos{\theta}\Rightarr ow x=\frac{\pi}{4}+\pi k, \ k\in\mathbb{Z}, \ \mbox{and} \ y=0$

and

$\displaystyle\cos{x}\sinh{y}=0 \Rightarrow x=\frac{\pi}{2}+\pi k, \ k\in\mathbb{Z}, \ \mbox{or when} \ y=-y\Rightarrow \ y=0$

However, looking at when $\displaystyle \cosh{y}=\cos{\theta}\Rightarrow \frac{e^y+e^{-y}}{2}=\frac{e^{\mathbf{i}\theta}+e^{-\mathbf{i}\theta}}{2}\Rightarrow e^y+e^{-y}=e^{\mathbf{i}\theta}+e^{-\mathbf{i}\theta}$

it appears that solution is only when $y=0$

When $\displaystyle x=\frac{\pi}{2}+\pi k, \ k\in\mathbb{Z}$

$\displaystyle \sin{\left(\frac{\pi}{2}+\pi k\right)}\cosh{y}=\cos{\theta}\Rightarrow (-1)^k\cosh{y}=\cos{\theta}\Rightarrow k=2p, \ p\in\mathbb{Z}$

$\displaystyle\Rightarrow\cosh{y}=\cos{\theta}\Righ tarrow y=\arccosh{(\cos{\theta})} \ \mbox{and} \ x=\frac{\pi}{2}+2\pi p$

One solution is: $\displaystyle \ \ \left(\frac{\pi}{2}+2\pi p,\cosh^{-1}{(\cos{\theta})}\right)$

When $y=0$

$\displaystyle \sin{x}\cosh{0}=\cos{\theta}\Rightarrow\sin{x}=\co s{\theta}\Rightarrow x=\frac{\pi}{4}+\pi m, \ m\in\mathbb{Z}$

Another solution is: $\displaystyle \ \ \left(\frac{\pi}{4}+\pi m, 0\right)$

$\displaystyle z=x+y\mathbf{i}\Rightarrow x=\frac{\pi}{4}+\pi m \ \mbox{when} \ y=0 \ \mbox{and} \ \ x=\frac{\pi}{2}+2\pi \ \mbox{when} \ y=\cosh^{-1}{(\cos{\theta})} \ m,p\in\mathbb{Z}$

Is this what can be said about z?

**Opalg** · December 23rd 2010, 12:40 AM

Originally Posted by dwsmith

If $\sin{z}=a, \ a\in [-1,1]$ , then what can be said about $z\mbox{?}$

Since a is between -1 and 1, $a=\cos{\theta}$ .

$\sin{z}=\sin{x}\cosh{y}+\mathbf{i}\cos{x}\sinh{y}$

If $z = x+iy$ then $\sin{z}=\sin{x}\cosh{y}+i\cos{x}\sinh{y}$ . If that number lies in the interval [–1,1] then its imaginary part must be 0. Therefore either $\sinh y = 0$ or $\cos x = 0$ .

If $\sinh y = 0$ then $y=0$ and hence $z$ is real. But every real number x satisfies $\sin x \in [-1,1]$ . So $z$ can be any real number.

Now suppose that $\sinh y \ne0$ and look at the other possibility, namely $\cos x = 0$ . If $\cos x = 0$ then $\sin x = \pm1$ ; and if $\sinh y \ne0$ then $\cosh y > 1$ . So in this case $|\sin{x}\cosh{y}| > 1$ and hence $\sin z\notin[-1,1]$ .

Conclusion: $\sin z \in [-1,1]\;\Longleftrightarrow\;z\in\mathbb{R}$ .

**dwsmith** · December 23rd 2010, 10:51 AM

Originally Posted by Opalg

If $z = x+iy$ then $\sin{z}=\sin{x}\cosh{y}+i\cos{x}\sinh{y}$ . If that number lies in the interval [–1,1] then its imaginary part must be 0. Therefore either $\sinh y = 0$ or $\cos x = 0$ .

If $\sinh y = 0$ then $y=0$ and hence $z$ is real. But every real number x satisfies $\sin x \in [-1,1]$ . So $z$ can be any real number.

Now suppose that $\sinh y \ne0$ and look at the other possibility, namely $\cos x = 0$ . If $\cos x = 0$ then $\sin x = \pm1$ ; and if $\sinh y \ne0$ then $\cosh y > 1$ . So in this case $|\sin{x}\cosh{y}| > 1$ and hence $\sin z\notin[-1,1]$ .

Conclusion: $\sin z \in [-1,1]\;\Longleftrightarrow\;z\in\mathbb{R}$ .

One question. By setting a = cos, I can eliminate the problem of |sinx coshy|>1 since I solved for the values that make it true. Does doing this make the approach wrong?

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