Equations of degree 2.

We solve the equation

$\displaystyle az^2+bz+c=0$    

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

We have:

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

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:

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

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:

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

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:

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

Noah Dana-Picard 2007-12-24