Hi I tried to solve the equation $$ z^2 - (1+m)(1+i)z + i(m^2+1)=0 $$ but I don't know if my answer is wrong or right.
My first $\Delta$ was $-2(im^2+i-2im)$, the second one $0$. So $$ z_1 = (1+m)(1+i)+i\sqrt{im^2+i-2im} $$ Am I right or wrong please?
By completing the square, we can obtain $$(z-(1+m)(1+i))^2=-\frac12 (5m^2+2m-1)~{\rm cis}~(\frac{3\pi}{2}+2k\pi)$$, where k is an integer. Then, we apply De'moivres theorem, which states that for some integer $n$, the set of solutions for $z^n=r\bigg(\cos(\theta+2k\pi) + i~\sin(\theta+k2\pi)\bigg)$ will be
$$\begin{align} % z_0 &= \sqrt[n]{r}\bigg(\cos(\theta/n) +i~\sin(\theta/n)\bigg) \\ % z_1 &= \sqrt[n]{r}\bigg(\cos(\frac{\theta}{n}+\frac{2\pi}{n}) +i~\sin(\frac{\theta}{n} +\frac{2\pi}{n}) \bigg) \\ % z_j &= \sqrt[n]{r}\bigg(\cos(\frac{\theta}{n}+\frac{2\pi j}{n}) +i~\sin(\frac{\theta}{n} +\frac{2\pi j}{n})\bigg) \text{ for all } j \leq n-1 \\ % \end{align}$$
So for your situation, $j=0,1$. Just do the computation and you will find the two answers in terms of $m$.
If someone finds a simpler method then please tell me.