# modulus of $\sin(z)$ where $z$ is a complex number

I'm asked to show if there exists $z$ in $\mathbb{C}$ such that, the two following conditions are simultaneously satisfied

$$|\sin(z)|>1, |\cos(z)|>1$$

For $|\sin(z)|^2$ I find $\displaystyle\frac{1-\cos(2z)}{2}$ wich is not a real number in general because $\cos(z)$ is not a real number. I don't know where is the mistake.

-

It is true that

$$\sin(z)^2 = \frac{1-\cos(2z)}{2}$$

but then you have to take the absolute value also on the right side. So:

$$|\sin(z)^2| = \left|\frac{1-\cos(2z)}{2}\right|$$

$$\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}, \cos(z) = \frac{e^{iz}+e^{-iz}}{2}$$

and try finding a value for $z$ such that the absolute value of both expressions is $>1$. Putting in purely imaginary numbers helps.

-
I mean modulus squared wich is $z\cdot z^*$ and the exercise that you propose is the actual exercise I'm trying to do. Moreover, i derived the expression in my question from the definitions of $\sin(z)$ and $\cos(z)$ I don't get the purely imaginary numbers part, because $e^{iz}=e^{i(x+iy)}=e^{ix-y}$ both real and imaginary part. – Jorge May 28 '12 at 16:51
So actually you're looking for $|\sin(z)|^2>1$ and $|\cos(z)|^2>1$? That does not make a difference, since $|z|>1$ if and only if $|z|^2>1$. Anyway, you can swap $|.|$ and $z^2$: $|z^2|=|z|^2$. Regarding the purely imaginary numbers, try setting $z=iy$ with $x=0$ and calculating $e^{iz}$ with that. – Gregor Bruns May 28 '12 at 17:24

Wouldn't it be enough just to pick an $x$ such that $e^{ix}$ is pretty big? Then since you have $\sin x\approx {\frac1{2i}e^{ix}}$ and $\cos x\approx \frac12 e^{ix}$ you would have what you wanted.

-
Hi, yes but I'm looking for an analytical expression, if possibl :) – Jorge May 28 '12 at 16:49
@burzum You said you were asked to show if there exists such a $z$. So it should be enough to produce one such $z$. What do you need exactly? – MJD May 28 '12 at 17:18
You're right indeed, my bad. I'm asked to determine a number $z$ such that the two conditions are satisfied. – Jorge May 28 '12 at 17:35