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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.

2 Answers 2

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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:

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

As for your exercise, write

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

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

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    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.2012-05-28
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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.


So for example, taking $x=-1000000000i$ ought to suffice.

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    You're right indeed, my bad. I'm asked to determine a number $z$ such that the two conditions are satisfied.2012-05-28