Quadratic Formula for $\cos\ z=2i$

Question

I saw this question on here and I have a question about it Solving the complex equation $\sin(z) = \cos(z)$

Below is how the person solved it:

$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$$ Then, $$\sin(z)=\cos(z)\implies \frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{iz}+e^{-iz}}{2}$$ Now let $e^{iz}=t$ ,then $e^{-iz}=1/t$ and you will get a quadratic equation,solve for it and back substitute it to get $z$.

The equation becomes $t^2=\frac{1+i}{1-i}=i=e^{i(\pi/2+2k\pi)}\implies t=e^{i(\pi/4+k\pi)},e^{i(5\pi/4+k\pi)}$. Equating it with $e^{iz}$ we get $z=\pi/4+k\pi$

Question: Can someone show/explain to me how $t^2 = (1/i)/(1-i)$?

Here is another example (first answer) how I don't understand $w^2 + 4ie + 1 = 0$ was even achieved.

Answer 1

It's was hard (for me) to understand, but I think it is:

$$\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{iz}+e^{-iz}}2\iff t-\frac1t=i\left(t+\frac1t\right)\iff t^2-1=it^2+i\iff$$

$$t^2(1-i)=1+i\iff t^2=\frac{1+i}{1-i}\;\;(=i\ldots)$$

Answer 2

$\displaystyle \frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{iz}+e^{-iz}}{2}\enspace $ => $\enspace \displaystyle e^{i2z}=\frac{1+i}{1-i}=(\frac{1+i}{\sqrt{2}})^2=e^{i\frac{\pi}{2}}$

=> $\enspace\displaystyle z=\frac{\pi}{4}+\pi k\enspace \in\mathbb{R}\enspace$ with $\enspace k\in\mathbb{Z}$

Answer 3

I'm not sure that I understand your problem. Maybe it's this? $$ t^1=\frac{1+i}{1-i}=(1+i)(1-i)^{-1}=(1+i)\frac{\overline{(1-i)}}{|1-i|^2}= $$ $$ =\frac{(1+i)(1+i)}{2}=\frac{1+i+i-1}{2}=i $$