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.

## marked as duplicate by Namaste, abiessu, Xander Henderson, Simply Beautiful Art, LeucippusJan 2 '18 at 2:28

• Do you mean that you tried the quadratic formula? – Toby Mak Jan 1 '18 at 23:07
• Well, you simply made computation mistakes with the quadratic formula, because it actually returns you $-1$ and $i$. – Crostul Jan 1 '18 at 23:08
• Thank you. I will check it again. – Chrysa Jan 1 '18 at 23:10
• $z^{2} + (1 - i) \cdot z - i = 0.\;$ Note that $a = 1, \; b = 1-i,\; c =- i$. Use the quadradic formula, $$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \iff z= \frac{(i-1 \pm \sqrt{ (1-i)^2 +4i} }{2}$$ and simplify. – Namaste Jan 1 '18 at 23:37
• I don't think this question, beyond the first two answers, needs additional answers. Overkill. – Namaste Jan 1 '18 at 23:46

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

• I see why $$2i = (i + 1)^2.$$ Is this a common trick when solving equations with complex numbers? – Chrysa Jan 1 '18 at 23:17
• @Chrysa This trick is evident when you compute the discriminant of your particular equation. First: you need to evaluate $(1-i)^2=-2i$; second: you need to find some number whose square is $2i$. Well, simply pick $i(1-i)$, i.e. $1+i$. In general, to compute a square root in $\Bbb C$ you have to switch to polar coordinates. – Crostul Jan 2 '18 at 9:40

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}

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.