Solving $\cos z = i$ for $z$
Solve $\cos z = i$ for $z$.
What I have tried: $$\cos z = i$$ $$\frac{e^{-zi}+e^{zi}}{2}=i$$ $$e^{-zi}+e^{zi}=2i=2e^{\frac\pi 2 + 2\pi k},\quad k\in \Bbb Z$$
I would take logs, but then I would have two terms in the left log, what now?
I have the solution for the problem already (this hasn't helped me since they have shown no work): $$-i\ln (\sqrt2 +1)+\left(2k+\frac12\right)\pi, \quad -i\ln(\sqrt 2 - 1) +\left(2k-\frac12\right)\pi, \qquad k\in \Bbb Z.$$
$\endgroup$ 35 Answers
$\begingroup$\begin{align} & e^{iz} + e^{-iz} = 2i \\ & e^{2iz} + 1 = 2ie^{iz} \qquad \text{(Both sides were multiplied by $e^{iz}$.)} \\ & w^2 + 1 = 2iw \\ & w^2 -2iw = -1 \\ & w^2 - 2iw - 1 = -2 \qquad\text{(completing the square)} \\ & (w- i)^2 = -2 \\ & w-i = \pm i\sqrt 2 \\ & w = i \pm i\sqrt 2 \\ & e^{iz} = i(1\pm\sqrt 2) = e^{i\pi/2} e^{\log(1\pm\sqrt 2)} \end{align} etc. (And notice that $\log(1-\sqrt 2) = -\log(1+\sqrt 2)$.)
$\endgroup$ 2 $\begingroup$You got to $e^{iz}+e^{-iz}=2i$. Let $u=e^{iz}$, this becomes:
$$u+\frac{1}{u}=2i$$ $$u^2+1=2iu$$ $$u^2-2iu+1=0$$ Thus $$u=\frac{2i\pm\sqrt{-4-4}}{2}$$
$$ u=\frac{2i\pm i\sqrt{8}}{2}$$ $$ u = i(1\pm\sqrt{2})$$
Then $$e^{iz}=i(1\pm\sqrt{2})$$
$\endgroup$ $\begingroup$It's convenient to use the decomposition of $z$ and $\cos z$ into real and imaginary parts, namely,$$\cos(x + iy) = \cos x \cosh y - i \sin x \sinh y$$(for $x, y \in \mathbb{R}$).
Using this formula to decompose $\cos z = i$ into real and imaginary parts gives the (equivalent) system\begin{align} \cos x \cosh y &= 0 \\ \sin x \sinh y &= 1 . \end{align}
$\endgroup$ $\begingroup$Now, $\cosh y > 0$ for all $y$, so the first equation is equivalent to $\cos x = 0$, which has solution $$x = \pi\left(m + \frac{1}{2}\right), \quad m \in \mathbb{Z}.$$ We can now substitute for $x$ in the other equation; in particular, $$\sin \pi\left(m + \frac{1}{2}\right) = (-1)^m,$$ so the second equation becomes $$(-1)^m \sinh y = 1,$$ and solving gives (recall here that $\sinh$ is injective, and so $\text{arsinh}$ is a bona fide inverse) $$y = \text{arsinh} [(-1)^m] = (-1)^m \ln(1 + \sqrt{2}).$$ So, the solutions of $\cos z = i$ are precisely $$z = \pi\left(m + \frac{1}{2}\right) + i (-1)^m \ln(1 + \sqrt{2}), \quad m \in \mathbb{Z}.$$ Specializing to $m = 2k$ and $m = 2k + 1$, together with the fact that $(\sqrt{2} + 1)(\sqrt{2} - 1) = 1$ and an appropriate logarithm identity, yields the given solutions.
When you have $e^{-iz} + e^{iz} = 2i$, multiply through with another factor of $e^{iz}$ to obtain $e^{2iz} -2ie^{iz} + 1 = 0$. Using the quadratic formula we get
$$ e^{iz} \;\; =\;\; \frac{2i \pm \sqrt{-4 - 4}}{2} \;\; =\;\; i(1 \pm \sqrt{2}). $$
The rest should be straightforward.
$\endgroup$ $\begingroup$Following the hint by @avid19:
Let $u=e^{iz}$
$$e^{-zi}+e^{zi}-2e^{\frac\pi 2 + 2\pi k}=0$$ $$\implies u^{-1}+u-2i=0$$ $$u^2-2iu+1=0$$
$$\text{by quadratic formula: } z = \frac{2i\pm\sqrt{-4-4}}{2}$$ $$u=e^{iz}=i\pm\frac{2\sqrt2 i}{2}=i\pm\sqrt{2}i$$ $$e^{iz}=i\pm \sqrt{2}i$$ $$iz = \ln(i\pm \sqrt{2}i)$$ $$z=\frac{\ln(i\pm \sqrt{2}i)}{i}=-i\ln(i\pm \sqrt{2}i)$$
As desired.
$\endgroup$
