Exercises — De Moivre's theorem — statement, proof, applications

What we do: apply the power rule directly. Why: the number is already in polar form (a length $\times$ a cis), so no conversion is needed — the theorem acts on length and angle separately.

• Modulus: $2^3=8$. Why cubed: each multiplication of the arrow by itself multiplies its length, so three multiplications give $2\times 2\times 2$.
• Angle: $3\times 30^{\circ }=90^{\circ }$. Why times 3: each multiplication adds the angle, so three of them add $30^{\circ }$ three times.

$z^3=8\mathrm{cis}(90^{\circ })=8(\cos 90^{\circ }+i\sin 90^{\circ })=8i.$ Answer: $8\mathrm{cis}(90^{\circ })=8i$.

Why no polar conversion is needed: the expression is already $\mathrm{cis}(20^{\circ })$ — an arrow of length $1$ at $20^{\circ }$ — so we can read off modulus and angle immediately.

• Modulus: $1^9=1$. Why it stays $1$: raising $1$ to any power is $1$; the arrow never changes length.
• Angle: $9\times 20^{\circ }=180^{\circ }$. Why times 9: angles add under multiplication, so nine copies of $20^{\circ }$ add to $180^{\circ }$.

$=\mathrm{cis}(180^{\circ })=\cos 180^{\circ }+i\sin 180^{\circ }=-1.$ Answer: $-1$.

Step 1 — polar form. Why convert at all? In Cartesian form $(1+i{)}^8$ would need eight messy multiplications; in polar form the power rule does it in one line. So we first rewrite the arrow as length $\times$ cis. $|1+i|=\sqrt{1^2+1^2}=\sqrt{2}$. Since real part $=$ imaginary part $=1$, the arrow sits on the $45^{\circ }$ line, i.e. $\pi /4$ radians. Why radians here: $\pi /4$ is a clean fraction of $\pi$, which multiplies neatly by $8$. Thus $1+i=\sqrt{2}\mathrm{cis}(\pi /4)$. Step 2 — apply De Moivre. Modulus $(\sqrt{2}{)}^8=2^4=16$ (why: the length is raised to the same power $8$); angle $8\times \pi /4=2\pi$ (why: angles add, so eight copies of $\pi /4$). $=16\mathrm{cis}(2\pi )=16(\cos 2\pi +i\sin 2\pi )=16.$ Why $\mathrm{cis}(2\pi )=1$: $2\pi$ radians is one full turn, landing the arrow back on the positive real axis. Answer: $16$.

Step 1 — recognise the arrow. Why check the modulus first? If the length is $1$ the number is a pure cis, and the power rule becomes trivial (length stays $1$). Modulus $=\sqrt{(1/2{)}^2+(\sqrt{3}/2{)}^2}=\sqrt{1/4+3/4}=1$. Reading the parts, $\cos =1/2$ and $\sin =\sqrt{3}/2$ is the angle $60^{\circ }$, so this is $\mathrm{cis}(60^{\circ })$. Step 2 — power. Modulus $1^6=1$ (why: $1$ to any power is $1$; the arrow keeps its length); angle $6\times 60^{\circ }=360^{\circ }$ (why: angles add, six copies of $60^{\circ }$). $=\mathrm{cis}(360^{\circ })=1.$ Why the result is $1$: $360^{\circ }$ is a full turn back to the start. Answer: $1$.

Step 1 — polar form. Why: a sixth power is painful in Cartesian form, so we convert to length $\times$ cis first. $|\sqrt{3}-i|=\sqrt{3+1}=2$. The arrow points into the fourth quadrant (positive real, negative imaginary), so its angle is $-30^{\circ }$. Why $-30^{\circ }$ and not $+30^{\circ }$: the imaginary part is negative, so the arrow dips below the real axis. Check: $2\cos (-30^{\circ })=2(\sqrt{3}/2)=\sqrt{3}$ ✔ and $2\sin (-30^{\circ })=2(-1/2)=-1$ ✔. So $\sqrt{3}-i=2\mathrm{cis}(-30^{\circ })$. Step 2. Modulus $2^6=64$ (why: length raised to the power $6$); angle $6\times (-30^{\circ })=-180^{\circ }$ (why: angles add, six copies of $-30^{\circ }$). $=64\mathrm{cis}(-180^{\circ })=64(\cos 180^{\circ }-i\sin 180^{\circ })=64(-1)=-64.$ Answer: $-64$.

Step 1 — polar form of the target. Why: the root formula needs the target written as $r\mathrm{cis}\varphi$. $8i=8\mathrm{cis}(90^{\circ })$: modulus $8$, angle $90^{\circ }$ (the arrow points straight up). Step 2 — roots formula with $n=3$, $r=8$, $\varphi =90^{\circ }$: $z_k=8^{1/3}\mathrm{cis}\left(\frac{90^{\circ }+360^{\circ }k}{3}\right)=2\mathrm{cis}(30^{\circ }+120^{\circ }k),k=0,1,2.$ Why $8^{1/3}=2$: to reach length $8$ after cubing, each root must have length $8^{1/3}=2$. Why $+120^{\circ }$ each step: a full turn ($360^{\circ }$) split into $n=3$ equal wedges, because adding $360^{\circ }$ to $\varphi$ names the same target $8i$ but a different root.

• $k=0$: $2\mathrm{cis}(30^{\circ })=2(\frac{\sqrt{3}}{2}+\frac{1}{2}i)=\sqrt{3}+i$.
• $k=1$: $2\mathrm{cis}(150^{\circ })=2(-\frac{\sqrt{3}}{2}+\frac{1}{2}i)=-\sqrt{3}+i$.
• $k=2$: $2\mathrm{cis}(270^{\circ })=2(0-i)=-2i$.

These sit on a circle of radius $2$, evenly $120^{\circ }$ apart (an equilateral triangle) — see the figure. Answer: $\sqrt{3}+i,-\sqrt{3}+i,-2i$.

Step 1 — polar form of the target. Why: the root formula only works once the right-hand side is $r\mathrm{cis}\varphi$. $-16$ is a negative real number, so its arrow points left: $-16=16\mathrm{cis}(180^{\circ })$, modulus $16$, angle $180^{\circ }$. Step 2 — roots formula with $n=4$, $r=16$, $\varphi =180^{\circ }$: $z_k=16^{1/4}\mathrm{cis}\left(\frac{180^{\circ }+360^{\circ }k}{4}\right)=2\mathrm{cis}(45^{\circ }+90^{\circ }k),k=0,1,2,3.$ Why $16^{1/4}=2$: each root must have length $2$ so that $2^4=16$ matches the target's length. Why $+90^{\circ }$ each step: $360^{\circ }$ split into $n=4$ equal wedges — each extra full turn on the target gives the next distinct root.

• $k=0$: $2\mathrm{cis}(45^{\circ })=\sqrt{2}+\sqrt{2}i$.
• $k=1$: $2\mathrm{cis}(135^{\circ })=-\sqrt{2}+\sqrt{2}i$.
• $k=2$: $2\mathrm{cis}(225^{\circ })=-\sqrt{2}-\sqrt{2}i$.
• $k=3$: $2\mathrm{cis}(315^{\circ })=\sqrt{2}-\sqrt{2}i$.

Four points, one in each quadrant, radius $2$, $90^{\circ }$ apart. Answer: $\pm \sqrt{2}\pm \sqrt{2}i$ (all four sign combinations).

Step 1 — two sides of one equation. By De Moivre, $(\cos \theta +i\sin \theta {)}^4=\cos 4\theta +i\sin 4\theta$. Expand the left with the Binomial theorem $(a+b{)}^4=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$, with $a=\cos \theta$, $b=i\sin \theta$. Write $c=\cos \theta$, $s=\sin \theta$: $c^4+4c^3(is)+6c^2(is{)}^2+4c(is{)}^3+(is{)}^4.$ Step 2 — powers of $i$: $i^2=-1$, $i^3=-i$, $i^4=1$: $=(c^4-6c^2{s}^2+s^4)+i(4c^{3}s-4c{s}^3).$ Step 3 — match real parts (two complex numbers are equal iff real and imaginary parts match): $\cos 4\theta =c^4-6c^2{s}^2+s^4.$ Step 4 — remove $s$. Substitute $s^2=1-c^2$: $=c^4-6c^2(1-c^2)+(1-c^2{)}^2=c^4-6c^2+6c^4+1-2c^2+c^4.$ $\cos 4\theta =8{\cos }^4\theta -8{\cos }^2\theta +1.$ Answer: $\cos 4\theta =8{\cos }^4\theta -8{\cos }^2\theta +1$.

Step 1. From $(\cos \theta +i\sin \theta {)}^5$, the imaginary part gives (binomial with $a=c$, $b=is$, using the odd-power terms $i^1,i^3,i^5$): $\sin 5\theta =5c^{4}s-10c^2{s}^3+s^5.$ Step 2 — divide by $s=\sin \theta$ (allowed since $s\ne 0$): $\frac{\sin 5\theta }{\sin \theta }=5c^4-10c^2{s}^2+s^4.$ Step 3 — substitute $s^2=1-c^2$: $=5c^4-10c^2(1-c^2)+(1-c^2{)}^2=5c^4-10c^2+10c^4+1-2c^2+c^4.$ $=16c^4-12c^2+1=16{\cos }^4\theta -12{\cos }^2\theta +1.■$

Step 1 — list the roots. The $n$-th roots of $1=\mathrm{cis}0$ are ${\omega }^k=\mathrm{cis}(2\pi k/n)$ for $k=0,\dots ,n-1$, where $\omega =\mathrm{cis}(2\pi /n)$. They are powers of one root $\omega$. Step 2 — sum them as a geometric series. With first term $1$, ratio $\omega$, $n$ terms: $S=1+\omega +{\omega }^2+\cdots +{\omega }^{n-1}=\frac{{\omega }^n-1}{\omega -1}.$ Why this formula: it is the standard sum of a geometric series, valid because $\omega \ne 1$ when $n\ge 2$. Step 3 — kill the numerator. By De Moivre, ${\omega }^n=\mathrm{cis}(2\pi )=1$, so ${\omega }^n-1=0$. Hence $S=0$. $■$ Verify $n=5$: the roots are $\mathrm{cis}(72^{\circ }k)$, $k=0,\dots ,4$. Real parts: $1+2\cos 72^{\circ }+2\cos 144^{\circ }$. Using $\cos 72^{\circ }=\frac{\sqrt{5}-1}{4}$ and $\cos 144^{\circ }=-\frac{\sqrt{5}+1}{4}$: $1+\frac{\sqrt{5}-1}{2}-\frac{\sqrt{5}+1}{2}=1+\frac{(\sqrt{5}-1)-(\sqrt{5}+1)}{2}=1+\frac{-2}{2}=0.✓$ Imaginary parts cancel by symmetry (each root and its conjugate pair up). Sum $=0$.

Step 1 — build a complex partner. Let $Z={\sum }_{k=0}^{n-1}\mathrm{cis}(k\theta )={\sum }_{k=0}^{n-1}(\cos k\theta +i\sin k\theta ).$ Then $C=\mathrm{Re}(Z)$ — the cosine sum is just the real part. Step 2 — geometric series with ratio $\mathrm{cis}\theta$ (using De Moivre, $\mathrm{cis}(k\theta )=(\mathrm{cis}\theta {)}^k$): $Z=\frac{(\mathrm{cis}\theta {)}^n-1}{\mathrm{cis}\theta -1}=\frac{\mathrm{cis}(n\theta )-1}{\mathrm{cis}\theta -1}(\theta \notin 2\pi \mathbb{Z}).$ Why the restriction $\theta \notin 2\pi \mathbb{Z}$: if $\theta$ is an integer multiple of $2\pi$ then $\mathrm{cis}\theta =1$, the denominator is $0$, and the formula breaks — that degenerate case is handled separately in the mistake box below. Step 3 — half-angle trick. Factor $\mathrm{cis}(n\theta )-1=\mathrm{cis}(n\theta /2)(\mathrm{cis}(n\theta /2)-\mathrm{cis}(-n\theta /2))=\mathrm{cis}(n\theta /2)\cdot 2i\sin (n\theta /2)$, and likewise the denominator gives $\mathrm{cis}(\theta /2)\cdot 2i\sin (\theta /2)$. Dividing: $Z=\mathrm{cis}\left(\frac{(n-1)\theta }{2}\right)\cdot \frac{\sin (n\theta /2)}{\sin (\theta /2)}.$ Step 4 — take the real part: $\boxed{C=\frac{\sin (n\theta /2)}{\sin (\theta /2)}\cos \left(\frac{(n-1)\theta }{2}\right).}$ Answer: the boxed Dirichlet-kernel formula.

Step 1 — polar form. Why: the root formula needs the target as $r\mathrm{cis}\varphi$. $|3-4i|=\sqrt{9+16}=5$. The arrow is in the fourth quadrant (positive real, negative imaginary), so $\varphi =-\arctan (4/3)$. Why arctan(4/3): for a fourth-quadrant arrow the angle below the axis has $\tan |\varphi |=\frac{\text{opposite}}{\text{adjacent}}=\frac{4}{3}$, giving $|\varphi |\approx 53.13^{\circ }$; the minus sign puts it below the real axis. Thus $3-4i=5\mathrm{cis}(-53.13^{\circ })$. Step 2 — square roots ($n=2$, $r=5$): $z_k=\sqrt{5}\mathrm{cis}\left(\frac{-53.13^{\circ }+360^{\circ }k}{2}\right),k=0,1.$

• $k=0$: angle $-26.57^{\circ }$, which is half of $-53.13^{\circ }$. Why the clean radicals $2/\sqrt{5}$ and $1/\sqrt{5}$: the half-angle $26.57^{\circ }$ satisfies $\tan 26.57^{\circ }=\frac{1}{2}$, so it is the angle of a right triangle with opposite $1$, adjacent $2$, hypotenuse $\sqrt{5}$. Hence $\cos 26.57^{\circ }=\frac{2}{\sqrt{5}}$ and $\sin 26.57^{\circ }=\frac{1}{\sqrt{5}}$. Therefore $z_0=\sqrt{5}(\frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}}i)=2-i.$
• $k=1$: angle $-26.57^{\circ }+180^{\circ }=153.43^{\circ }$, which flips the arrow to its opposite: $z_1=-(2-i)=-2+i.$

Check: $(2-i{)}^2=4-4i+i^2=3-4i$ ✔. Answer: $z=\pm (2-i)$.