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solve the equation $\cos{z}=\sqrt{2}$ for z

i am making a mistake somewhere, here is my attempt $$z=-i\log({\sqrt{2}+i(1-2)^{1/2})}$$$$z=-i\log(\sqrt{2}-1)$$$$z=-i[\ln(\sqrt{2}-1)+i(2n\pi)]$$$$z=2n\pi + -i\ln(\sqrt{2}-1)$$

2 Answers2

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$$ \cos(x+iy) = \cos x\cosh y - i\sin x\sinh y$$

so we have $$ \cos x \cosh y = \sqrt{2} \\ \sin x \sinh y = 0$$

If $\sinh y = 0$ then $\cosh y = 1$ and $\cos x = \sqrt{2}$ which yields no solution, so $\sin x = 0$

Furthermore, $\cosh y \ge 0$, so $\cos x \ge 0$ or $\cos x = 1$ which means $x = 2n\pi$

This leaves $\cosh y = \sqrt{2}$ or $y = \pm\ln (\sqrt{2} + 1)$, so the solution is

$$ z = 2n\pi \pm i \ln (\sqrt{2} + 1) $$

Dylan
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  • You might at least note that your $ z = 2n\pi \pm i \ln (\sqrt{2} + 1) $ is equivalent to the OP's $ z = 2n\pi \pm i \ln (\sqrt{2} - 1) $... – Did Mar 23 '18 at 06:06
  • I think OP meant to write $2n\pi - i\ln(\sqrt{2}-1)$ which would make an incomplete answer. – Dylan Mar 23 '18 at 06:56
  • Even if it was supposed to be $\pm$, there's no way they would have arrived at both signs, given the work before it. – Dylan Mar 23 '18 at 06:59
  • Yes the question is a mess, but it does write $+-$. And anyway, this is not my main point here, which is that you ought to describe precisely the relationship of your solution with what the OP did (or did not do). – Did Mar 23 '18 at 07:00
  • Fair enough. I din't do that when I first wrote the answer, because I believed OP's wasn't correct, so it wouldn't matter. My answer wasn't picked anyway. Thanks for pointing it out in the comments so I don't have to. – Dylan Mar 23 '18 at 07:05
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I am not quite following you in your work. Here is a set up: With $cos^2z+sin^2z=1$ we find $sinz=\pm{i}$. Now use identity $e^{iz}=cosz+isinz$ with the given info of $cosz$ and $sinz$. One solution would be $e^{iz}=\sqrt{2}+1$ from which you can take the $ln$ to find $iz=ln(\sqrt{2}+1)$ and that you can solve for $z$ by multiplying both sides by $-i$. Can you find the other solution? Also beware of the multiplicities.

imranfat
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  • my book defines $$arccos(z)=-i\log(z+i\sqrt{1-z^{2}})$$ –  Nov 08 '17 at 03:22
  • But is that formula helpful in this question? You still need $z$. I usually use Euler's identity for these problems. Goes very easy – imranfat Nov 08 '17 at 03:27
  • This might be helpful: https://math.stackexchange.com/questions/1166562/solve-sinz-2?rq=1 – imranfat Nov 08 '17 at 03:28
  • This is perhaps better. Sine goes same way of course: https://math.stackexchange.com/questions/1945602/finding-z-in-mathbb-c-with-sin-z-2/1945713#1945713 – imranfat Nov 08 '17 at 03:29
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    My professor has only solved these types of problems using the definition i posted, but i do see how yours makes sense. do you see any errors with my original post? –  Nov 08 '17 at 03:37
  • Well, $ (1-2)^\frac{1}{2}$ is essentially $\sqrt{-1}$ but in this context, that ought to be $i$ and $-i$. yielding two solutions. You only got one – imranfat Nov 08 '17 at 03:42
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    then would the full solution be $-i\ln(\sqrt{2} \pm 1)$ –  Nov 08 '17 at 03:56
  • I think that's what I also would get from my approach (multiplicities too, complex cosine is still periodic) – imranfat Nov 08 '17 at 04:03
  • ok thank you for the help –  Nov 08 '17 at 04:05