Solving $\cos z=i\sin z$, $z$ complex number.
Well, for me is not solution because after replace $$\frac{e^{iz}+e^{-iz}}{2}=i\frac{e^{iz}-e^{-iz}}{2i}$$ equivalently
$$2e^{-iz}=0$$
but here i dont find solution. Is fine my attempt?
Thank you
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2Looks right to me. – saulspatz Sep 08 '21 at 14:33
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Yes you are right, indeed by $z=x+iy$ we have
$$|e^{-iz}|=|e^{-ix+y}|=|e^y||e^{-ix}|=e^y>0$$
or as an alternative pointed out by MartinR
$$e^{iz}e^{-iz}=e^0=1$$
therefore no solution exists.
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1Better (IMO): $e^{-iz} e^{iz} = 1$. I would always calculate directly with complex numbers if possible, instead of resorting real and imaginary parts. – Martin R Sep 08 '21 at 15:12
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@MartinR Thanks! Indeed this is really fine. Of course this is a very useful approach, more elegant, I agree with that. – user Sep 08 '21 at 15:18