Solution of quadratic equations with real coefficients and a complex variable.

Solve the quadratic equation.

1. $z^2+2z+5=0.$

Solution.

$D=2^2-4\cdot 1\cdot 5=4-20=-16$

$z_1=\frac{-2+\sqrt{-16}}{2}=\frac{-2+\sqrt{16}\sqrt{-1}}{2}=\frac{-2+4i}{2}=-1+2i;$

$z_2=\frac{-2-\sqrt{-16}}{2}=\frac{-2-\sqrt{16}\sqrt{-1}}{2}=\frac{-2-4i}{2}=-1-2i.$

Answer: $z_1=-1+2i;$ $z_2=-1-2i.$

2. $4z^2-2z+1=0.$

Solution.

$D=(-2{)}^2-4\cdot 4\cdot 1=4-16=-12$

$z_1=\frac{2+\sqrt{-12}}{8}=\frac{2+\sqrt{12}\sqrt{-1}}{8}=\frac{2+2\sqrt{3}i}{8}=\frac{1}{4}+\frac{\sqrt{3}}{4}i;$

$z_2=\frac{2-\sqrt{-12}}{8}=\frac{2-\sqrt{12}\sqrt{-1}}{8}=\frac{2-2\sqrt{3}i}{8}=\frac{1}{4}-\frac{\sqrt{3}}{4}i.$

Answer: $z_1=\frac{1}{4}+\frac{\sqrt{3}}{4}i;$ $z_2=\frac{1}{4}-\frac{\sqrt{3}}{4}i.$

Solve the biquadratic equation

3. $z^4+18z^2+81=0.$

Solution.

Let's make a variable substitution:

$t=z^2.$

We get a quadratic equation:

$t^2+18t+81=0.$

Let's solve it:

$D=(18{)}^2-4\cdot 1\cdot 81=324-324=0.$

$t_1=\frac{-18+0}{2}=-9;$

$t_2=\frac{-18-0}{2}=-9.$

Then let's make the inverse substitution:

$t_1=t_2=z_{1,2}^2=z_{3,4}^2\Rightarrow$

$\Rightarrow -9=z_{1,2}^2\Rightarrow$

$\Rightarrow z_{1,2}=\pm \sqrt{-9}=\pm \sqrt{9}\sqrt{-1}=\pm 3i.$

Answer: $z_{1,2}=z_{3,4}=\pm 3i.$

Homework

Solve the biquadratic equation

1. $z^4+4z^2+3=0.$

Answer: $z_{1,2}=\pm i$ $z_{3,4}=\pm \sqrt{3}i.$

2. $z^4+9z^2+20=0.$

Answer: $z_{1,2}=\pm 2i$ $z_{3,4}=\pm \sqrt{5}i.$

Tags: complex analysis, complex numbers, quadratic equations