Find the solutions of the equation $z^2+2z+1-i=0$ in the form$ z=p+iq$

I had tried this so far : let $z = x + iy$ Hence, $$(x + iy)^2 + 2 (x + iy) + 1 - i = 0$$ $$x^2 + 2xyi + - y^2 + 2x + 2iy + 1 - i = 0$$ $$x^2 - y^2 + 2x + 1 + 2xyi + 2iy - i = 0$$

It's way easier to keep the complex variable whole

$$z^2+2z+1=(z+1)^2=i \implies z = -1\pm\sqrt{i} = -1\pm\frac{1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}}$$

Gotta get used to the fact that complex numbers are actually numbers in their own right.

$z^2 +2z + 1-i=0$ will have solutions by the quadratic formula of

$z = \frac {-2 \pm \sqrt {4 - 4(1-i)}}2$

So factoring out the common factor $2$ we get $z = -1\pm i^{\frac 12}=$

$-1 \pm (\frac 1{\sqrt 2} + \frac 1{\sqrt 2}i) =$

$(-1\pm \frac 1{\sqrt 2}) + i(\pm \frac 1{\sqrt 2})$.

....

Although you could continue.

You must solve $x^2 - y^2 + 2x + 1=0$ and $2xy + 2y-1 = 0$.

So $y = x^2 +2x + 1$ so $y = (x+1)^2$ and $y = \pm(x+1)$.

But we also have $2xy + 2y - 1 = 0$ so $y(2x+2) = 1$ and $y = \frac 1{2(x+1)}$.

So we must solve $\pm (x+1) = \frac 1{2(x+1)}$ or

$\pm (x+1)^2 = \frac 12$ so $(x+1)^2 = \frac 12$ (and $y > 0$) and $x+1 =\pm \frac 1{\sqrt 2}$ and ... the rest falls out.

$x = -1 \pm \frac 1{\sqrt 2}$ and $y = \pm \frac 1{\sqrt 2}$.

(I actually wasn't expecting it to fall out so neatly.)