<s>[INST] Solve $\cos(z)=\frac{3}{4}+\frac{i}{4}$ I tried solving this using the definition of $cos(z)=\frac{e^{iz}+e^{-iz}}{2}$ and equating it to $\frac{3}{4}+\frac{i}{4}$ and converting it to a complex quadratic equation through a substitution $t=e^{iz}$ and finding roots via the complex quadratic formula but it didn't seem to work. I would prefer solutions via elementary methods.
Here is my attempt: 
By definition we have $\frac{e^{iz}+e^{-iz}}{2}=\frac{3}{4}+\frac{i}{4} \implies e^{iz}+e^{-iz}=\frac{3}{2}+\frac{i}{2}$. Let $t=e^{iz}$ so then we have $t+\frac{1}{t}=\frac{3}{2}+\frac{i}{2}$ and if we multiply both sides by $t$ we have $t^2+1=(\frac{3}{2}+\frac{i}{2})t$ and hence $t^2+(\frac{3}{2}+\frac{i}{2})t+1=0$ By the quadratic formula for complex numbers we have, $a=1, b=\frac{3}{2}+\frac{i}{2}, c=1 \implies z=\frac{-(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2})^2-4(1)(1)}}{2(1)}$. Simplyifing we have $z=\frac{-\frac{3}{2}-(\frac{1}{2})i \pm \sqrt{-2+(\frac{3}{2})i}}{2}$ We wish to express $-2+(\frac{3}{3})i$ in polar form so we have $|-2+(\frac{3}{2})i|=\frac{5}{2}$. Now equating the real and imaginary parts we have $\frac{5}{2}\cos(\theta)=-2 \implies \cos(\theta)=-\frac{4}{5}$ and $\frac{5}{2}\sin(\theta)=\frac{3}{2} \implies \sin(\theta)=\frac{3}{5}$. From this we have $\tan(\theta)=-\frac{3}{4} \implies \theta=\arctan(-\frac{3}{4}) \approx -.6435$ rad. So we have $w=-2+(\frac{3}{2})i=\frac{5}{4}(\cos(\theta)+i\sin(\theta))=\frac{5}{4}e^{i\theta}$. By Proposition 1.3.12 we have $\sqrt{w}=\sqrt{\frac{5}{4}}e^{\frac{i\theta}{2}}=\frac{\sqrt{5}}{2}e^{\frac{i\theta}{2}}$. Similarily for $-\frac{3}{2}-(\frac{1}{2})=\frac{\sqrt{10}}{2}e^{i\varphi}$ Where $\varphi=\arctan(\frac{1}{3})$.  So finally we have $z=\frac{-(\frac{3}{2}+\frac{i}{2}) \pm \sqrt{(\frac{3}{2}+\frac{i}{2})^2-4(1)(1)}}{2(1)}=\frac{\sqrt{10}e^{i\varphi} \pm \sqrt{5}e^{\frac{i\theta}{2}}}{4}$ as solutions to $\cos(z)=\frac{3}{4}+\frac{i}{4}$.
 [/INST] One possible "elementary method" might start:
$$\cos(x+iy) = \cos(x)\cos(iy)-\sin(x)\sin(iy),$$
which simplifies to
$$\cos(x)\cosh(y)-i\sin(x)\sinh(y).$$
Now let
$$\cos(x)\cosh(y)-i\sin(x)\sinh(y) \equiv \frac{3}{4}+\frac{i}{4},$$
which gives $$\cos(x)\cosh(y)=3/4$$ and $$\sin(x)\sinh(y)=-1/4.$$
You may be able to solve for $x$ and $y$.
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