De Moivre's theorem

Complex numbers, grade 12 (5 units) Lesson 11 of 18Learning

De Moivre's theorem

The theorem

A power is repeated multiplication, and each multiplication adds the angles. So in $\displaystyle z^n$ the modulus is raised to the power $\displaystyle n$ , and the angle is multiplied by $\displaystyle n$ :

\displaystyle \left[R(\cos\varphi+i\sin\varphi)\right]^{n}

\displaystyle =

\displaystyle R^{n}(\cos n\varphi+i\sin n\varphi)

The theorem holds for every natural $\displaystyle n$ , and also for $\displaystyle {n}={0}$ and for negative integers, since $\displaystyle {z^{-n}}={\frac{1}{z^n}}$ .

A proof by induction

The proof below proves the theorem for every natural $\displaystyle n$ by induction, with the product rule in polar form:

Base: for $\displaystyle {n}={1}$ both sides are $\displaystyle R\operatorname{cis}\varphi$ .
Assumption: suppose that for some natural $\displaystyle k$ , $\displaystyle {(R\operatorname{cis}\varphi)^k}={R^k\operatorname{cis}k\varphi}$ .
Step: $\displaystyle {(R\operatorname{cis}\varphi)^{k+1}}={(R\operatorname{cis}\varphi)^k\cdot R\operatorname{cis}\varphi}={R^k\operatorname{cis}k\varphi\cdot R\operatorname{cis}\varphi}$ .
By the product rule: $\displaystyle {}={R^{k+1}\operatorname{cis}(k\varphi+\varphi)}={R^{k+1}\operatorname{cis}(k+1)\varphi}$ – the statement for $\displaystyle k+1$ .
Conclusion: by the principle of induction, the theorem holds for every natural $\displaystyle n$ .

Worked example: a high power

To raise a number in algebraic form to a high power, move to the polar form. The example below works out $\displaystyle (1-i)^{10}$ :

$\displaystyle {1-i}={\sqrt{2}\operatorname{cis}\left(-\frac{\pi}{4}\right)}$ .
$\displaystyle {(1-i)^{10}}={(\sqrt{2})^{10}\operatorname{cis}\left(-\frac{10\pi}{4}\right)}={32\operatorname{cis}\left(-\frac{5\pi}{2}\right)}$ .
$\displaystyle {-\frac{5\pi}{2}+2\pi}={-\frac{\pi}{2}}$ , so $\displaystyle {(1-i)^{10}}={32\operatorname{cis}\left(-\frac{\pi}{2}\right)}={-32i}$ .

Double-angle formulas

De Moivre’s theorem gives $\displaystyle {(\cos\theta+i\sin\theta)^2}={\cos2\theta+i\sin2\theta}$ . Opening the brackets on the left side:

$\displaystyle {(\cos\theta+i\sin\theta)^2}={\cos^2\theta-\sin^2\theta+2i\sin\theta\cos\theta}$ .
Comparing the real parts: $\displaystyle {\cos2\theta}={\cos^2\theta-\sin^2\theta}$ .
Comparing the imaginary parts: $\displaystyle {\sin2\theta}={2\sin\theta\cos\theta}$ .

Exercise 1

Calculate with De Moivre’s theorem, and write the result in algebraic form.

/ – 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

Exercise 2

Calculate, and write the result in polar form.

/ – 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 3

Open the brackets in $\displaystyle (\cos\theta+i\sin\theta)^3$ , and write the result in the form $\displaystyle \ldots+i\cdot\ldots$ .
Then compare with the parts of $\displaystyle \cos3\theta+i\sin3\theta$ , and find a formula for $\displaystyle \cos3\theta$ in terms of $\displaystyle \cos\theta$ and $\displaystyle \sin\theta$ .

/ – 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

Find the smallest natural number $\displaystyle n$ for which $\displaystyle (1+\sqrt{3}i)^n$ is real, and the smallest natural number for which it is purely imaginary – if there is one. Give reasons by the angle.

Real

/ – 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 5

The sequence $\displaystyle {z_n}={(\cos\alpha+i\sin\alpha)^n}$ is defined for every natural $\displaystyle n$ , with $\displaystyle 0<\alpha<\pi$ .

Show that $\displaystyle |z_{n+1}-z_n|$ does not depend on $\displaystyle n$ : take $\displaystyle z_n$ out as a common factor.

What does the result mean in the Gauss plane?

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