Let $x\in (0,\frac{\pi}{2})$ and let $z\in \mathbb{C}$
We consider the equation $z^2-2iz-(1+e^{i4x})=0$
Write the two solutions of the equation in exponential form.
What I tried
$\delta=(-2i)^2+4+4e^{i4x}=(2e^{i2x})^2$
$x_1=\frac{2i-e^{i2x}}{2}=i-e^{i2x}=e^{i \frac{\pi}{2}}-e^{i2x}$
$x_2=\frac{2i+e^{i2x}}{2}=i+e^{i2x}=e^{i \frac{\pi}{2}}+e^{i2x}$
How can I write $x_1$ and $x_2$ in the form $re^{iy}$ with $y\in (-\pi , \pi)$?
The roots of the equation are $$ i+e^{2ix}\qquad\text{and}\qquad i-e^{2ix} $$ Let's consider the first one recalling $i=e^{i\pi/2}$: $$ i+e^{2ix}=i(1+e^{-i\pi/2}e^{2ix})=i(1+e^{i(2x-\pi/2)}) $$ Set $2x-\pi/2=2y$, for simplicity,: $$ i+e^{2ix}= i(1+e^{2iy})=e^{i\pi/2}e^{iy}(e^{iy}+e^{-iy})= 2\cos y\cdot e^{i(y+\pi/2)} $$ Returning in terms of $x$, $y=x-\pi/4$, so $y+\pi/2=x+\pi/4$ and so $$ i+e^{2ix}=2\cos\left(x-\frac{\pi}{4}\right)\cdot e^{i(x+\pi/4)} $$
You can try the other root.
The trick is to get the number in the form $e^{2i\varphi}+1$ or $e^{2i\varphi}-1$ so the technique of collecting $e^{i\varphi}$ will leave something more manageable.
$\Delta=(-2i)^2+4+4e^{i4x}=(2e^{i2x})^2$
$$e^{ia}=cos a+i sin a$$ $$x_1=\frac{2i-e^{i2x}}{2}=i-e^{i2x}=i-(cos2x+i \sin 2x)=\\-cos 2x +i(1-sin2x) \to |x_1|=\sqrt{cos^22x+(1-sin2x)^2}\\tan \theta=\frac{-cos 2x}{1-sin2x}=\frac{-(cos ^x-sin^2x)}{sin^2x+cos^2x-2sinxcosx}=\frac{-(cos x-sinx)(cosx+sinx)}{(cosx-sinx)^2}=\\\frac{-(cosx+sinx)}{(cosx-sinx)}=\\\frac{-\frac{(cosx+sinx)}{cosx}}{\frac{(cosx-sinx)}{cosx}}=\frac{-(tanx+1)}{1-tanx}=-tan(\frac{\pi}{4}+x)=tan(-\frac{\pi}{4}-x)$$so $$\large x_1=\frac{2i-e^{i2x}}{2}=\sqrt{cos^22x+(1-sin2x)^2}e^{(-\frac{\pi}{4}-x)}$$ then you can do like this for $x_2$ $$x_2=\frac{2i+e^{i2x}}{2}=i+e^{i2x}=i+cos 2x+i sin2x$$