$z^2+\left(1-i\right)z+\left(i-1\right)=0$
My answer: $\frac{1}{2}\left(-1+i\pm \left(\sqrt{\sqrt{13}+2}-\frac{1}{3}\sqrt{\sqrt{13}-2}i\right)\right)$
Textbook answer $\frac{1}{2}\left(-1+i\pm \left(\sqrt{\sqrt{13}+2}-\sqrt{\sqrt{13}-2}i\right)\right)$
My troubles begin with the square root $\sqrt{4-6i} = x+iy$
I actually was able to get the textbook answer by considering purely $x^2 + y^2$ and $x^2 - y^2$ BUT I am unable to get the same answer using the $2xy = -6$ to find $y$ because I always get a $\frac{1}{3}$ after rationalizing. I have actually always done it this way: a mixture of considering $x^2 - y^2 = 4 \tag{1}$ and the silmultanoues equations formed by $x^2 +y^2 = 2\sqrt{13} \tag{2}$, then putting $x$ into $2xy = -6$. For this particular question, I always get $\frac{1}{2}\left(-1+i\pm \left(\sqrt{\sqrt{13}+2}-\frac{1}{3}\sqrt{\sqrt{13}-2}i\right)\right)$
When I go the correct answer, it was by adding AND substracting $[1]$ and $[2]$ to determine $x$ and $y$ separately. Why wont the mother method work though? Is it because rationalising with nested square root and the $\pm$ and complex number $i$ all at the same time gets wierd/messy/undefined?
