The equation is $z^2 = 1 + i$. I solved this equation by replacing z by x + iy and then simplifying everything that can be. Then I got my z1 & z2 by using a system. However, although my solution seems correct, in the handbook they solved the problem using $e$ and got the following solution: $$ 2^{1/4} e^{i \frac{\pi}{8}} \quad \text{and} \quad 2^{1/4} e^{i \frac{9\pi}{8}} $$
It isn't explained anywhere, so I'm kinda stuck on how they solved it this way. If anyone could point me in the right direction I would appreciate it.
Welcome!
They simply used the exponential form of a complex number: any $z=x+iy\in\mathbf C$ can be written as $$z=r\mathrm e^{i\theta},\quad\text{ where }\; r=|z|\;\text{ and }\;\mathrm e^{i\theta}=\cos\theta+i\sin\theta.$$ The argument of $z$, $\theta$, is the polar angle of the image of $z$ in the Argand-Cauchy plane, defined modulo $2\pi$.
We have $\;1+i=\sqrt2\,\mathrm e^{\tfrac{i\pi}4}$, so the equation $z^2=1+i$ is equivalent to $$r^2\mathrm e^{2i\theta}=\sqrt2\,\mathrm e^{\tfrac{i\pi}4}\iff\begin{cases}r^2=\sqrt 2,\\2\theta\equiv\frac\pi 4\mod 2\pi. \end{cases}$$