For which values of $a$ lie the roots of $z^2+2a(1+i)z+(4+2a^2i)=0$ in the first quadrant of complex plane?

Question

Given is$$z^2+2a(1+i)z+(4+2a^2i)=0$$ I'm trying to solve it for z by completing the square but I'm stuck. For it to be in the first quadrant both the real and the imaginary part should be positive if I'm not mistaken.

I wrote it as $2ix^2+(2+2i)xz+z^2=-4$ and now I want to square the left side, since the root of the right side is $\pm2i$.

Answer 1

You can write the equation as$$z^2+2a(1+i)z+(4+2a^2i)=\Big(z+a(1+i)\Big)^2+4=0$$where we have used $\Big((1+i)a\Big)^2=a^2(1+i)^2=2ia^2$ therefore$$z_1=2i-a(1+i)=(2-a)i-a\\z_2=-2i-a(1+i)=(-2-a)i-a$$and they fall in the first quadrant if $$2-a>0\\-a>0\\-2-a>0$$which yield to $$a<2\\a<0\\a<-2$$

Conlusion

For $a<-2$, the roots of $z^2+2a(1+i)z+(4+2a^2i)=0$ reside in the first quadrant.