Roots of order n

Complex numbers, grade 12 (5 units) Lesson 14 of 18Learning

Roots of order n

The solutions of

\displaystyle {z^n}={w}

As with the roots of unity, substitute $\displaystyle {z}={r\operatorname{cis}\theta}$ in the equation $\displaystyle {z^n}={R\operatorname{cis}\varphi}$ :

$\displaystyle {r^n\operatorname{cis}n\theta}={R\operatorname{cis}\varphi}$ , so $\displaystyle {r^n}={R}$ and $\displaystyle {n\theta}={\varphi+2k\pi}$ .
$\displaystyle {r}={\sqrt[n]{R}}$ and $\displaystyle {\theta}={\frac{\varphi}{n}+\frac{2k\pi}{n}}$ .

\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 k=0,1,\ldots,n-1

The equation has $\displaystyle n$ different solutions, all on a circle centred at the origin with radius $\displaystyle \sqrt[n]{R}$ .
The angle between two neighbouring solutions is $\displaystyle \frac{2\pi}{n}$ : they are the vertices of a regular polygon, turned by the angle $\displaystyle \frac{\varphi}{n}$ .

Worked example:

\displaystyle {z^3}={8i}

The example below solves $\displaystyle {z^3}={8i}$ :

$\displaystyle {8i}={8\operatorname{cis}\frac{\pi}{2}}$ : $\displaystyle {R}={8}$ , $\displaystyle {\varphi}={\frac{\pi}{2}}$ , $\displaystyle {n}={3}$ .
$\displaystyle {\sqrt[3]{8}}={2}$ , and $\displaystyle {\theta}={\frac{\pi}{6}+\frac{2k\pi}{3}}$ .
$\displaystyle {k}={0}$ : $\displaystyle {2\operatorname{cis}\frac{\pi}{6}}={\sqrt{3}+i}$ .
$\displaystyle {k}={1}$ : $\displaystyle {2\operatorname{cis}\frac{5\pi}{6}}={-\sqrt{3}+i}$ .
$\displaystyle {k}={2}$ : $\displaystyle {2\operatorname{cis}\frac{3\pi}{2}}={-2i}$ .

The three solutions are the vertices of an equilateral triangle in the circle of radius 2, as in the coordinate system below.

\displaystyle x^{2}+y^{2}=4

\displaystyle z_0=(1.732,1)

\displaystyle z_1=(-1.732,1)

\displaystyle z_2=(0,-2)

From one root to all the roots

If $\displaystyle z_0$ is a solution of $\displaystyle {z^n}={w}$ , all the solutions are $\displaystyle z_0$ times the roots of unity of order $\displaystyle n$ : $\displaystyle {(z_0\omega^k)^n}={z_0^n\cdot(\omega^n)^k}={w}$ . In the Gauss plane – $\displaystyle z_0$ turned again and again by the angle $\displaystyle \frac{2\pi}{n}$ .

$\displaystyle 1+i$ is a solution of $\displaystyle {z^4}={w}$ , so $\displaystyle {w}={(1+i)^4}={-4}$ .
The roots of unity of order 4 are $\displaystyle 1$ , $\displaystyle i$ , $\displaystyle -1$ , $\displaystyle -i$ , so the solutions are $\displaystyle 1+i$ , $\displaystyle -1+i$ , $\displaystyle -1-i$ , $\displaystyle 1-i$ .

Exercise 1

Solve. Write each solution in algebraic or polar form, the angles in radians.

Exercise 2

$\displaystyle \sqrt{3}-i$ is a solution of the equation $\displaystyle {z^6}={w}$ .

Find $\displaystyle w$ .

/ – 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

Find the other solutions.

Exercise 3

The solutions of $\displaystyle {z^6}={64}$ are the vertices of a regular hexagon.
Calculate the area of the hexagon: it is made of six equilateral triangles whose side is the radius of the circle.

/ – 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

Exercise 4

The solutions of $\displaystyle {(z-2i)^4}={81}$ are the vertices of a square.
Find the centre of the square and its vertices: substitute $\displaystyle {u}={z-2i}$ .

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 $\displaystyle z_0$
$\displaystyle w=z_{0}+(z-z_{0})\operatorname{cis}\theta$