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How would you go about finding all the solutions to $\cos(z)=0$, where $z\in \mathbb{C}$?

I have $$\cos(z)=0 \implies (e^{iz})^2=-1 \implies \text{Log}(e^{iz})=\text{Log}(e^{i\pi(1/2+2k)})\qquad\text{ where }k\in \mathbb{Z}$$

$$\implies ie^{-y}x=i\pi(1/2+2k)$$

$$\implies z= e^{y}\pi(1/2+2k)+iy \qquad\forall k\in \mathbb{Z}\;\text{ and }\;\forall y\in \mathbb{R}$$

Is this correct?

Thanks!

EDIT:

$e^{2iz}=e^{i\pi(1+2k)} \Rightarrow z=\pi/2+k\pi$ where $k\in \mathbb{Z}$

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  • $\begingroup$ In short, no. It turns out that the only complex zeroes are the same as the real zeroes. Plug in any value for your y that's not zero, and you'll get a non-zero cosine. $\endgroup$
    – davidlowryduda
    Commented Oct 23, 2011 at 16:28
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    $\begingroup$ The roots of the cosine are all real. One way to establish this is to consider the real and imaginary parts of $\cos(x+iy)$ (hint: use the addition formula and the relation between trigonometric and hyperbolic functions.) $\endgroup$ Commented Oct 23, 2011 at 16:28

3 Answers 3

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Logarithms are treacherous. It is safer and easier to note that we are looking at the equation $e^{iz}=-e^{-iz}$. Let $z=a+ib$, where $a$ and $b$ are real. Then the norm of $e^{i(a+ib)}$ is $e^{-b}$, and the norm of $-e^{-iz}$ is $e^{b}$. If $b \ne 0$, the norms don't match.

So $z$ has to be real, and we are in familiar territory.

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No. The part where you seem to be assuming $\log\mathrm e^{\mathrm iz}=\mathrm i\mathrm e^{-y}x$ is wrong. Rather, (the principal value of) that logarithm is $\log\mathrm e^{\mathrm iz}=\mathrm iz$. (The natural logarithm is the inverse of the exponential function.)

You also got a factor of $2$ wrong in the arguments somewhere, presumably in trying to resolve the square. The possible arguments for numbers whose square is $-1$ are $\pi\left(\frac12+k\right)$, without the $2$.

Then by comparing real and imaginary parts you can conclude that $x=\pi\left(\frac12+k\right)$, $y=0$.

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  • $\begingroup$ Yes, I realize now I messed up with my definition of log, I've fixed it with my edit, thanks! that's very helpful $\endgroup$
    – Freeman
    Commented Oct 23, 2011 at 16:38
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Since you know that $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$. So $\frac{e^{iz} + e^{-iz}}{2}=0$ thus ${e^{iz} + e^{-iz}} = 0$ suppose you let $u = e^{iz}$. You now have $u + \frac{1}{u} = 0$, then $u^2 +1 =0$; hence $(u-i) (u+i) = 0$. Therefore $u=i$ or $u=-i$.

We now substitute back to get $e^{iz}=i$ or $e^{iz}=-i$; by taking natural log from both side we see that $e^{iz}=-i$ can not be a solution thus $e^{iz}=i$ is a solution. Hence $z= -i\ln(i)$. So the value that satisfied this equation is when $z= -i\ln(i).$

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  • $\begingroup$ Can't you simply say that since $e^{iz} = i $ or $-i$ then $z = \pi + 2k\pi$ with $k \in \mathbb{Z}$ $\endgroup$ Commented Jan 8, 2017 at 17:05

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