A question in my complex analysis book (Gamelin's "Complex Analysis", question I.8.7) asks me to prove that $e^{iz} = \cos(z) \pm \sqrt{\cos^2(z) - 1}$. Using the identity $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$ and the quadratic formula this is easy enough:

$\begin{align*} 2\cos(z) &= e^{iz} + e^{-iz} \\ 2\cos(z) e^{iz} &= e^{2iz} + 1 \\ 0 &= e^{2iz} - 2\cos(z) e^{iz} + 1 \\ e^{iz} &= \frac{2\cos(z) \pm \sqrt{4\cos^2(z) - 4}}{2} \\ e^{iz} &= \cos(z) \pm \sqrt{\cos^2(z) - 1} \end{align*}$

But now I am trying to interpret this result.

  • Does this imply that $e^{iz}$ is a multi-valued function, one value for the $+$ branch and one value for the $-$ branch?
  • Since $\cos$ is an even function (i.e. $\cos(z) = \cos(-z)$ for all $z$), $$\cos(z) \pm \sqrt{\cos^2(z) - 1} = \cos(-z) \pm \sqrt{\cos^2(-z) - 1},$$ correct? But $e^{iz} \neq e^{-iz}$ in general.

Help on either or both bullets would be appreciated.

  • 2
    $\begingroup$ Whether you take $+$ or $-$ just depends on which branch of the square root you take, and I suspect the ambiguity is just to make it clear that you do need to take care in the branch cut. But this is a terrible question as it's written, for just that reason - the LHS is entire, while the RHS must have a cut somewhere. $\endgroup$ Apr 14, 2015 at 0:14
  • 1
    $\begingroup$ Ok. For the curious, the question as stated in the book reads "Set $w = \cos(z)$ and $\zeta = e^{iz}$. Show that $\zeta = w \pm \sqrt{w^2 - 1}$..." $\endgroup$
    – kbrose
    Apr 14, 2015 at 0:18
  • $\begingroup$ That is indeed what I did in my proof, just without the substitutions $\zeta = e^{iz}$ and $w = \cos(z)$. $\endgroup$
    – kbrose
    Apr 14, 2015 at 1:07


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