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My attempt:
$$\cos z = \frac{e^{iz} + e^{-iz}}{2} = 2 \\e^{2iz} - 4e^{iz} + 1 = 0\\ e^{iz} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}.$$
Since the RHS is real, the complex logarithm is equal to the real logarithm, so
$$iz = \ln (2\pm \sqrt{3})$$

Is this correct? I have no solutions so I'd like to confirm. How do I account for periodicity here?

Twenty-six colours
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1 Answers1

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You're on the right track. I would do two additional things.

1) When you take natural logarithm of complex numbers an additive constant of $i2k\pi$ comes in, as the logarithm is multivalued. Thus

$iz=\ln (2\pm \sqrt 3)+i2k\pi$

2) Multiply by $-i$ to isolate $z$. Note that the numbers $\ln (2\pm \sqrt 3)$ are negatives of each other:

$z=-i \ln (2\pm \sqrt 3)+2k\pi=i\ln (2\mp \sqrt 3)+2k\pi$

Oscar Lanzi
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  • Thanks! I wasn't sure about the additive constants for some reason. Is it possible for some result to give an odd number of $\pi$? I did a similar question, using same reasoning as you did; but the solution had $(2k+1)\pi$.
    The original question is solving $\cosh z = -5$.     – Twenty-six colours Aug 02 '17 at 12:05
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    Indeed you can get an odd number times $\pi$. The inverse cosine of a negative number $<-1$ does that. – Oscar Lanzi Aug 02 '17 at 12:10
  • Hmm, is there a place which shows this? Quite new and want to get the basics down first. I tried solving the quadratic $u^2 + 10u + 1 = 0$ where $u = e^{iz}$, and using the same method, I would have added a multiple of $2\pi i$ – Twenty-six colours Aug 02 '17 at 12:12
  • You do indeed add multiples of $2\pi$ for $z$ (meaning, multiples of $i2\pi$ for the full exponent) in your equation. But all the possible values fold back into one pair of roots when you compute $u=\exp(iz)$. – Oscar Lanzi Jul 09 '18 at 20:13