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I'm given $$\cos(z) = \frac{5}{2}$$ and I'm trying to solve for $z$ but I keep going in circles. I know $\cos z = 5/2 = 1/2(e^{iz}+e^{-iz})$ so then $e^{iz}+e^{-iz} = 5$ but then I'm stuck

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Taking $t=e^{iz}$ we get $$t+\frac{1}{t}=5 \implies t^2-5t+1=0 \implies t_{1,2} = \frac{5 \pm \sqrt{21}}{2}$$

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Note

$$\cos z = \cosh(-i z +i 2\pi n)=\frac52 $$

which leads to $-i z +i 2\pi n=\pm \cosh^{-1}\frac52$ and the solutions

$$z= 2\pi n\pm i \cosh^{-1}\frac52$$

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Solving the quadratic further

$$ z= \pm i \log\;(\dfrac{5 + \sqrt{21}}{2})$$ which is pure imaginary to which we add the real variable part $ 2 \pi n $ making up the complex angle.

Graphs of $ (\cos x, \frac52) $ do not intersect. It is interesting to note however that the real part corresponds to the closest points between the non-intersecting cosine curve and straight line.

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