Equations of degree 2.

We solve the equation

$$az^2+bz+c=0$$

where $a,b,c$ are complex numbers and $a \neq 0$ .

We have:

$$\forall z \in \mathbb{C}, az^2+bz+c = a \left[ \left( z+ \frac {b}{2a} \right) ^2 + \frac {b^2-4ac}{4a^2} \right]$$

Define $\Delta = b^2-4ac$ .

1. If $\Delta =0$ , then $az^2+bz+c = 0 \Leftrightarrow a \left( z+ \frac {b}{2a} \right) ^2 =0$ . The equation has a double solution, namely $z= -\frac {b}{2a}$ .
2. $\Delta \neq 0$ , then $\Delta$ has two complex square roots; we denote them $\delta$ and $-\delta$ . Thus:

   $\forall z \in \mathbb{C}, az^2+bz+c = a \left( z+ \frac {b+\delta}{2a} \right) \left( b + \frac {b+\delta}{2a} \right)$ .

   The equation has two distinct complex solutions, namely:

$$z_1 = \frac {-b-\delta}{2a} \; \; z_1 = \frac {-b+\delta}{2a}$$

Example 1.6.1   Solve the equation $z^2-iz+1=0$ , where $z$ is a complex unknown.

$\Delta = (-i)^2-4 \cdot 1 \cdot 1 = -5$ . Thus, $\Delta$ has two complex square roots, namely $i \sqrt{5}$ and $-i \sqrt{5}$ . The solutions $z_1$ and $z_2$ of the equation are given by:

$$z_1=\frac {i + i \sqrt{5}}{2}= \frac {1+ \sqrt{5}}{2} i \quad z_2=\frac {i - i \sqrt{5}}{2}= \frac {1- \sqrt{5}}{2} i.$$

Example 1.6.2   Solve the equation $z^2-(1+3i)z-2+2i=0$ , where $z$ is a complex unknown.

$\Delta = (1+3i)^2+4(2-2i)=-2i$ . The complex square roots of $\Delta$ are $1-i$ and $-1+i$ (You can compute them either by the algebraic method, described in subsection 5.3, or by the trigonometric method, described in subsection 5.1).

The solutions $z_1$ and $z_2$ of the equation are given by:

$$z_1=\frac {1+3i + 1 -i}{2}=1+i \quad z_2= \frac {1+3i - 1+i}{2}= 2i.$$