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Here's what I have so far:

Let $e^z + e^{-z} = 0$. Then $e^x\cos y + ie^x\sin y + e^{-x}\cos y - ie^{-x}\sin y = 0$.

The simplification in the second term follows because I know $\cos(y) = \cos(-y)$ and $\sin(y) = -\sin(y)$.

Am I going about this the right way?

Not totally sure how to simplify from here...

Thanks for the help, Mariogs

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  • $\begingroup$ Yes, you are going the right way. Group the terms in the expression into real and imaginary parts, and ask yourself when a complex number is equal to $0$. $\endgroup$ – Lucian Mar 9 '14 at 15:04
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...This happens exactly when $\mathrm e^{2z}+1=0$ (why?), that is, when $\mathrm e^{2z}=-1=\mathrm e^{\mathrm i\pi}$, that is, when $2z=\mathrm i\pi+2n\mathrm i\pi$ for some integer $n$ (why?), which is equivalent to the fact that $z=\mathrm i\pi(n+\frac12)$ for some integer $n$.

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    $\begingroup$ BTW, $e^z + e^{-z} = 2\cosh z$, so these are exactly the zeros of the hyperbolic cosine (in the complex plane). $\endgroup$ – fgp Mar 9 '14 at 15:06
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Multiply by $e^z$; so you have $e^{2z}+1=0$ and then $z=\pm \frac{i \pi }{2}$ from which you can conclude that ...

I am sure that you can take from here.

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  • $\begingroup$ "and then $z=\pm \frac{i \pi }{2}$" Actually, no. $\endgroup$ – Did Mar 9 '14 at 15:46
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You may going forward from your solution.
$$e^x\cos y + ie^x\sin y + e^{-x}\cos y - ie^{-x}\sin y = 0$$ That means, that $$e^x\cos y+e^{-x}\cos y=0$$ and $$ie^x\sin y- ie^{-x}\sin y = 0$$ because complex number is equal to zero when its real and imaginary parts are equal to zero. Mind, that $x$ and $y$ are real.

$(e^x+e^{-x})\cos y=0$ means $\cos y =0$ because $(e^x+e^{-x})$ is always positive. And $\cos y =0$ means $y=\frac{\pi}{2}+\pi n$

If $\cos y =0$ then $\sin y \neq0$ and from $$ie^x\sin y- ie^{-x}\sin y = 0$$ we can obtain $$e^x- e^{-x}=0$$ It has one solution: $x=0$.

So, finally, $z=x+iy=i\left(\frac{\pi}{2}+\pi n\right)$.

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