Solutions of quadratic equation with complex coefficients, using the quadradic formula.

Question

I have a hard time solving the following equation.

$$z^{2} + (1 - i) \cdot z - i = 0$$

I tried factorization and got $z = -1$ or $z = i$, which I suppose are correct.

However, when I try the quadratic formula for solving second degree equations, the results I get don't match with the solutions I get above.

Answer 1

$$b^2-4ac = (1-i)^2 - 4(-i) = -2i+4i= 2i = (i+1)^2$$ Hence the two solutions are $$z_1= \frac{-1+i+(i+1)}{2}=i; \ \ \ \ z_2 = \frac{-1+i-(i+1)}{2}=-1$$

Answer 2

Useful facts $$\sqrt{x+yi}=\pm \left( \sqrt{\frac{\sqrt{x^2+y^2}+x}{2}} +\frac{iy}{|y|}\sqrt{\frac{\sqrt{x^2+y^2}-x}{2}} \right)$$

\begin{align} z &= \frac{i-1 \pm \sqrt{(1-i)^2+4i}}{2} \\ &= \frac{i-1 \pm \sqrt{2i}}{2} \\ &= \frac{i-1 \pm (1+i)}{2} \\ &= -1 \quad \text{or} \quad i \end{align}

Refer to another answer here

Answer 3

Just sorting for the equations on the real and imaginary coordinates, $$ x^2 - y^2 + x + y = 0 \\ 2xy + y - x - 1 = 0 $$ we get these possible solutions:

The intersection points $(-1, 0) = -1$ and $(0, 1) = i$ give the solutions.