Solving a complex trigonometric equation

Question

$$\cos(Z)=\frac{3i}{4}$$ I solved it in a weird way by setting $$\cos(Z) = \cos(x+iy) = \cos(x)\cosh(y)-i \sin(x)\sinh(y)$$ and then setting this equal $$\frac{3}{4}i$$ , and since the cosine function has no zero values, then cosine x has to be equal to zero hence $$x= \frac{\pi}{2}+ 2 \pi n $$ $n= 0, 1, 2,\ldots$, and the same for the imaginary part setting $$\sinh(y)=\frac{3}{4}$$ The problem is I need help solving it the normal way setting $$\cos(Z)=\frac{e^{iz}+e^{-iz}}{2}$$

Answer 1

The "normal" way is by setting $cos(z)$ as you defined, then defining $e^{iz}=t$. Multiply by t and you'll get a solvable equation (second degree polynomial in t).

Answer 2

$$\frac{e^{iz}+e^{-iz}}{2}=\frac{3i}{4}$$ $$\frac{(e^{iz})^2+1}{2e^{iz}}=\frac{3i}{4}$$ $$2(e^{iz})^2-3ie^{iz}+2=0$$ Solving above quadratic equation for $e^{iz}$ $$e^{iz}=\frac{-(-3i)\pm\sqrt{(-3i)^2-4\cdot 2\cdot 2}}{2\cdot 2}$$ $$e^{iz}=2i, -\frac{i}{2}$$ You can proceed by setting $e^{iz}=e^{-y+ix}=e^{-y}e^{ix}=e^{-y}(\cos x+i\sin x)$