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If I have $\cos(z)=2$ I can say $\cos(a+ib)=2$

using double angle ideas $\cos(a)\cos(ib)+\sin(a)\sin(ib)=2$ using Euler's formula $\cos(a)\cosh(b)+i\sin(a)\sinh(b)=2$ equating real and imaginary parts $\cos(a)\cosh(b)=2$, $\sin(a)\sinh(b)=0$.

from here I'm unsure how to solve this set of simultaneous equations.

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  • $\begingroup$ Do you have to use the trig, hyperbolig trig formulae? I think there's a more direct way. $\endgroup$ – Simon S Apr 26 '15 at 16:45
  • $\begingroup$ I'm not sure but if I say that z=arccos(2) I end up with a purely real solution which makes me think that It's missing something? $\endgroup$ – Goods Apr 26 '15 at 16:48
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    $\begingroup$ If $\arccos 2$ is real, you have something wrong with your $\arccos$.. $\endgroup$ – GEdgar Apr 26 '15 at 16:50
  • $\begingroup$ hmmm.. yes it actually gives an error. I must have done cos before... $\endgroup$ – Goods Apr 26 '15 at 16:52
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Hint: A better way to do this. $\cos z = (e^{iz}+e^{-iz})/2$. Then your equation $\cos z = 2$ becomes a quadratic equation for $e^{iz}$.

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  • $\begingroup$ so e^i2z-4e^iz+1=o and solve that? $\endgroup$ – Goods Apr 26 '15 at 16:50
  • $\begingroup$ ...so if you actually want real and imaginary parts in your answer, you will need to know how to do real and imaginary parts of the complex logarithm. $\endgroup$ – GEdgar Apr 26 '15 at 17:05
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Alternatively, as $\cos z = {1 \over 2} (e^{iz} + e^{-iz})$, setting this equal to $2$, and writing $w = e^{iz}$, we have

$${1 \over 2} (w + 1/w) = 2 \ \ \text{ or alternatively } w^2 + 1 = 4w$$

$w = 2 \pm \sqrt 3$. Now solve for $z$.

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$$e^{iz}=\cos z+i\sin z=2+i\cdot\pm\sqrt{1-2^2}=2\pm\sqrt3$$

$$\implies iz=\ln(2\pm\sqrt3)\iff z=-i\ln(2\pm\sqrt3)$$

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