Square roots and quadratic equations

Complex numbers, grade 12 (5 units) Lesson 6 of 18Learning

Square roots and quadratic equations

The square roots of a complex number

Every complex number $\displaystyle w\neq0$ has two square roots – two solutions of the equation $\displaystyle {z^2}={w}$ , opposites of each other. You find them by substituting $\displaystyle {z}={a+bi}$ . The example below solves $\displaystyle {z^2}={5-12i}$ :

$\displaystyle {(a+bi)^2}={a^2-b^2+2abi}$ , so $\displaystyle {a^2-b^2}={5}$ and $\displaystyle {2ab}={-12}$ .
The moduli are equal too: $\displaystyle {|z|^2}={|w|}$ , that is $\displaystyle {a^2+b^2}={\sqrt{25+144}}={13}$ .
Adding the first and the third equation: $\displaystyle {2a^2}={18}$ , so $\displaystyle {a}={\pm3}$ .
From $\displaystyle {2ab}={-12}$ : when $\displaystyle {a}={3}$ , $\displaystyle {b}={-2}$ ; when $\displaystyle {a}={-3}$ , $\displaystyle {b}={2}$ .
The roots: $\displaystyle {z}={3-2i}$ or $\displaystyle {z}={-3+2i}$ .

A quadratic equation with real coefficients

The quadratic formula holds in the complex numbers too. When the coefficients are real and the discriminant is negative, $\displaystyle {\sqrt{\Delta}}={i\sqrt{-\Delta}}$ , and the solutions are two conjugate numbers. The example below solves $\displaystyle {z^2+2z+5}={0}$ :

$\displaystyle {\Delta}={4-20}={-16}$ , and $\displaystyle {\sqrt{-16}}={4i}$ .
$\displaystyle {z}={\frac{-2\pm4i}{2}}$ , that is $\displaystyle {z}={-1+2i}$ or $\displaystyle {z}={-1-2i}$ .

So in the complex numbers every quadratic equation with real coefficients has two solutions: two real solutions (perhaps equal), or two conjugate solutions.

Worked example: complex coefficients

When the coefficients are complex, the discriminant is complex too, and you find its root as at the start of the lesson. The example below solves $\displaystyle {z^2-(3+i)z+4+3i}={0}$ :

$\displaystyle {\Delta}={(3+i)^2-4(4+3i)}={8+6i-16-12i}={-8-6i}$ .
A root of $\displaystyle -8-6i$ : $\displaystyle {a^2-b^2}={-8}$ , $\displaystyle {2ab}={-6}$ , $\displaystyle {a^2+b^2}={10}$ , so $\displaystyle {a}={\pm1}$ and $\displaystyle {\sqrt{\Delta}}={1-3i}$ (or its opposite).
$\displaystyle {z}={\frac{3+i\pm(1-3i)}{2}}$ , that is $\displaystyle {z}={2-i}$ or $\displaystyle {z}={1+2i}$ .

Here the solutions are not conjugates: the solutions come in conjugate pairs only when the coefficients are real.

An equation from its solutions

If $\displaystyle z_1$ and $\displaystyle z_2$ are the solutions of $\displaystyle {z^2+bz+c}={0}$ , then $\displaystyle {z^2+bz+c}={(z-z_1)(z-z_2)}$ , and opening the brackets:

\displaystyle z_{1}+z_{2}

\displaystyle =

\displaystyle -b

\displaystyle z_{1}\cdot z_{2}

\displaystyle =

\displaystyle c

The example below builds an equation with real coefficients that has $\displaystyle 1+i$ as one of its solutions:

The coefficients are real, so $\displaystyle 1-i$ is a solution too.
$\displaystyle {z_1+z_2}={2}$ and $\displaystyle {z_1z_2}={1+1}={2}$ .
The equation: $\displaystyle {z^2-2z+2}={0}$ .

Exercise 1

Find the square roots of $\displaystyle -8+6i$ .
Substitute $\displaystyle {z}={a+bi}$ in $\displaystyle {z^2}={-8+6i}$ : solve the system in $\displaystyle a$ and $\displaystyle b$ .

Exercise 2

Solve.

Exercise 3

Solve with the quadratic formula.

Exercise 4

$\displaystyle 3-2i$ is a solution of the equation $\displaystyle {z^2+bz+c}={0}$ , with $\displaystyle b$ and $\displaystyle c$ real.
Find $\displaystyle b$ , $\displaystyle c$ and the other solution.

\displaystyle b

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle c

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

The other solution

Exercise 5

$\displaystyle 2+i$ is a solution of the equation $\displaystyle {z^2-(3+2i)z+c}={0}$ .
Find $\displaystyle c$ and the other solution: use the sum of the solutions.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: complex numbers, grade 12 (5 units)

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The algebraic form and the four operations

The number $\displaystyle i$
$\displaystyle i^{2}=-1$
The algebraic form
$\displaystyle z=a+bi$ $\displaystyle a=\operatorname{Re}(z),\ b=\operatorname{Im}(z)$
Equality of complex numbers
$\displaystyle a+bi=c+di\iff a=c,\ b=d$
Multiplying
$\displaystyle (a+bi)(c+di)$ $\displaystyle =$ $\displaystyle (ac-bd)+(ad+bc)i$
The conjugate
$\displaystyle \bar{z}=a-bi$
A number times its conjugate
$\displaystyle z\cdot\bar{z}$ $\displaystyle =$ $\displaystyle a^{2}+b^{2}$
Dividing – times the denominator’s conjugate
$\displaystyle \frac{z_{1}}{z_{2}}$ $\displaystyle =$ $\displaystyle \frac{z_{1}\cdot\bar{z}_{2}}{z_{2}\cdot\bar{z}_{2}}$

The modulus and the conjugate

The modulus – the distance from the origin
$\displaystyle |z|$ $\displaystyle =$ $\displaystyle \sqrt{a^{2}+b^{2}}$
The distance between two points
$\displaystyle |z_{1}-z_{2}|$ $\displaystyle =$ $\displaystyle \sqrt{(a_{1}-a_{2})^{2}+(b_{1}-b_{2})^{2}}$
The product with the conjugate and the modulus
$\displaystyle z\cdot\bar{z}$ $\displaystyle =$ $\displaystyle |z|^{2}$
The modulus of a product
$\displaystyle |z_{1}\cdot z_{2}|$ $\displaystyle =$ $\displaystyle |z_{1}|\cdot|z_{2}|$
The conjugate of a sum
$\displaystyle \overline{z_{1}+z_{2}}$ $\displaystyle =$ $\displaystyle \bar{z}_{1}+\bar{z}_{2}$
The conjugate of a product
$\displaystyle \overline{z_{1}\cdot z_{2}}$ $\displaystyle =$ $\displaystyle \bar{z}_{1}\cdot\bar{z}_{2}$
A number plus its conjugate
$\displaystyle z+\bar{z}$ $\displaystyle =$ $\displaystyle 2\operatorname{Re}(z)$
The triangle inequality
$\displaystyle |z_{1}+z_{2}|\le|z_{1}|+|z_{2}|$

The quadratic equation

The quadratic formula
$\displaystyle z_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
The sum of the roots
$\displaystyle z_{1}+z_{2}$ $\displaystyle =$ $\displaystyle -\frac{b}{a}$
The product of the roots
$\displaystyle z_{1}\cdot z_{2}$ $\displaystyle =$ $\displaystyle \frac{c}{a}$

The polar form (angles in radians)

The polar form
$\displaystyle z=r(\cos\theta+i\sin\theta)$ $\displaystyle r=|z|$
The notation cis
$\displaystyle r\operatorname{cis}\theta$ $\displaystyle =$ $\displaystyle r(\cos\theta+i\sin\theta)$
The real part
$\displaystyle a=r\cos\theta$
The imaginary part
$\displaystyle b=r\sin\theta$
The angle – by the quadrant of the point
$\displaystyle \tan\theta=\frac{b}{a}$ $\displaystyle a\neq0$
A product in polar form
$\displaystyle z_{1}\cdot z_{2}=r_{1}\cdot r_{2}\left[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})\right]$
A quotient in polar form
$\displaystyle \frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}\operatorname{cis}(\theta_{1}-\theta_{2})$
De Moivre’s theorem
$\displaystyle \left[R(\cos\varphi+i\sin\varphi)\right]^{n}$ $\displaystyle =$ $\displaystyle R^{n}(\cos n\varphi+i\sin n\varphi)$
The solutions of $\displaystyle {z^n}={R(\cos\varphi+i\sin\varphi)}$
$\displaystyle z_{k}=\sqrt[n]{R}\left[\cos\left(\frac{\varphi}{n}+\frac{2k\pi}{n}\right)+i\sin\left(\frac{\varphi}{n}+\frac{2k\pi}{n}\right)\right]$ $\displaystyle z^{n}=R(\cos\varphi+i\sin\varphi),\ k=0,1,\ldots,n-1$

Loci and rotation

The circle with centre $\displaystyle z_0$ and radius $\displaystyle R$
$\displaystyle |z-z_{0}|=R$
The perpendicular bisector of the segment $\displaystyle z_1z_2$
$\displaystyle |z-z_{1}|=|z-z_{2}|$
Apollonius’ circle
$\displaystyle |z-z_{1}|=m|z-z_{2}|$ $\displaystyle m>0,\ m\neq1$
Rotating by $\displaystyle \theta$ about $\displaystyle z_0$
$\displaystyle w=z_{0}+(z-z_{0})\operatorname{cis}\theta$