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Year 12 Specialist (Unit 3 & 4) Further complex numbers

Complex arithmetic in polar form (De Moivre)

20 practice questions 0 video lessons Theory + worked examples

Master complex arithmetic in polar form for Year 12 Specialist Mathematics in Queensland (QCAA). De Moivre’s theorem raises a complex number in modulus–argument form to any integer power by raising the modulus and multiplying the argument, turning long expansions into a single step.

You will learn to apply De Moivre’s theorem for positive and negative powers, reduce answers to the principal argument, and prove the modulus and argument identities — the polar toolkit that leads on to the roots of complex numbers and factorising over the complex field.

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Theory

Complex arithmetic in polar form uses De Moivre’s theorem to raise a complex number to a power in Year 12 Specialist Mathematics (QCAA, Queensland). Writing \(z=r\operatorname{cis}\theta\), a power just raises the modulus and multiplies the argument: \((r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)\). This page also proves the modulus and argument identities that make polar multiplication so quick, worked in full.

The polar (modulus–argument) form of a complex number is \(z=r\operatorname{cis}\theta\), where \(r=\lvert z\rvert\ge 0\) is the modulus, \(\theta=\arg z\) is the argument, and \(\operatorname{cis}\theta=\cos\theta+i\sin\theta\). By convention the principal argument satisfies \(-\pi<\theta\le\pi\).

De Moivre’s theorem states that for every integer \(n\), \((r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)\). To raise \(z\) to a power, raise the modulus to that power and multiply the argument by \(n\). It holds for negative integers too, so \(z^{-2}=r^{-2}\operatorname{cis}(-2\theta)\).

De Moivre delivers two modulus–argument identities at once: \(\lvert z^n\rvert=\lvert z\rvert^n\) and \(\arg(z^n)=n\arg z\). Two more come from the polar rule for a product: \(\lvert z_1 z_2\rvert=\lvert z_1\rvert\lvert z_2\rvert\) and \(\arg(z_1 z_2)=\arg z_1+\arg z_2\) — moduli multiply and arguments add. The conjugate identity \(z\bar z=\lvert z\rvert^2\) gives \(\lvert z\rvert\) without surds.

Because multiplying the argument by \(n\) can push it outside \((-\pi,\pi]\), the last step is often to reduce to the principal argument by adding or subtracting a multiple of \(2\pi\).

Powers of a complex number spiralling outward on an Argand diagram For z equal to root two cis pi over four, the points z, z squared and z cubed are plotted as vectors from the origin. Each power multiplies the modulus by root two and adds pi over four to the argument, so the points spiral outward with equal angle steps. Re Im z z² z³
Powers of \(z=\sqrt{2}\operatorname{cis}\dfrac{\pi}{4}\): each step multiplies the modulus by \(\sqrt{2}\) and adds \(\dfrac{\pi}{4}\), so the powers spiral outward.
The product of two complex numbers on an Argand diagram Vectors for z1 with modulus two at thirty degrees, z2 with modulus one point five at sixty degrees, and their product z1 z2 with modulus three at ninety degrees. The moduli multiply, two times one point five equals three, and the arguments add, thirty plus sixty equals ninety degrees. Re Im z₁ z₂ z₁z₂
Product on the Argand plane: moduli multiply \((2\times1.5=3)\) and arguments add \((30^\circ+60^\circ=90^\circ)\), so \(z_1 z_2=3\operatorname{cis}\dfrac{\pi}{2}\).

De Moivre’s theorem for any integer \(n\) (raise the modulus, multiply the argument):

\[ (r\operatorname{cis}\theta)^n = r^n\operatorname{cis}(n\theta) \]
(rcisθ)n=rncis(nθ)

Modulus and argument of a power:

\[ \lvert z^n\rvert = \lvert z\rvert^{\,n}, \qquad \arg(z^n) = n\arg z \]
|zn|=|z|n

Product identities (moduli multiply, arguments add):

\[ \lvert z_1 z_2\rvert = \lvert z_1\rvert\lvert z_2\rvert, \qquad \arg(z_1 z_2) = \arg z_1 + \arg z_2 \]
|z1z2|=|z1||z2|

Conjugate identity (a real, non-negative number):

\[ z\bar z = \lvert z\rvert^2 \]
zz¯=|z|2
Principal argument. After multiplying by \(n\), the angle \(n\theta\) may fall outside \((-\pi,\pi]\). Add or subtract a multiple of \(2\pi\) so the final argument satisfies \(-\pi<\theta\le\pi\).

How to evaluate a power with De Moivre’s theorem

  1. Write \(z\) in polar form: find the modulus \(r=\lvert z\rvert\) and the argument \(\theta=\arg z\), so \(z=r\operatorname{cis}\theta\).
  2. Apply De Moivre: raise the modulus to the power, \(r^n\), and multiply the argument, \(n\theta\).
  3. Reduce to the principal argument: if \(n\theta\) is outside \((-\pi,\pi]\), add or subtract a multiple of \(2\pi\).
  4. Convert back if asked: expand \(r^n(\cos(n\theta)+i\sin(n\theta))\) to give the answer in Cartesian form \(a+bi\), keeping surds exact.
Example 1 — Power in polar form
Use De Moivre’s theorem to evaluate \(\left(3\operatorname{cis}\dfrac{\pi}{6}\right)^{3}\), giving your answer in polar form.
Solution

Raise the modulus to the power and multiply the argument by \(n=3\):

\(r^n\)\(=\)\(3^{3}\)
\(=\)\(27\)
\(n\theta\)\(=\)\(3\times\dfrac{\pi}{6}\)
\(=\)\(\dfrac{\pi}{2}\)
\(\left(3\operatorname{cis}\dfrac{\pi}{6}\right)^{3}\)\(=\)\(27\operatorname{cis}\dfrac{\pi}{2}\)

\(\left(3\operatorname{cis}\dfrac{\pi}{6}\right)^{3}=27\operatorname{cis}\dfrac{\pi}{2}\).

Example 2 — Convert, power, then back to \(a+bi\)
Evaluate \((1+\sqrt{3}\,i)^{5}\) by first writing \(1+\sqrt{3}\,i\) in polar form, giving your answer in Cartesian form \(a+bi\).
Solution

Find the modulus and argument (first quadrant):

\(\lvert z\rvert\)\(=\)\(\sqrt{1^2+(\sqrt{3})^2}\)
\(=\)\(\sqrt{4}=2\)
\(\arg z\)\(=\)\(\dfrac{\pi}{3}\)
\(z\)\(=\)\(2\operatorname{cis}\dfrac{\pi}{3}\)

Apply De Moivre, reduce to the principal argument, then convert:

\(z^{5}\)\(=\)\(2^{5}\operatorname{cis}\left(5\times\dfrac{\pi}{3}\right)\)
\(=\)\(32\operatorname{cis}\dfrac{5\pi}{3}\)
\(\dfrac{5\pi}{3}-2\pi\)\(=\)\(-\dfrac{\pi}{3}\)
\(z^{5}\)\(=\)\(32\left(\cos\left(-\tfrac{\pi}{3}\right)+i\sin\left(-\tfrac{\pi}{3}\right)\right)\)
\(=\)\(32\left(\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i\right)\)
\(=\)\(16-16\sqrt{3}\,i\)

\((1+\sqrt{3}\,i)^{5}=16-16\sqrt{3}\,i\).

Example 3 — Negative integer power
Use De Moivre’s theorem to evaluate \(\left(2\operatorname{cis}\dfrac{\pi}{3}\right)^{-2}\), giving your answer in polar form.
Solution

De Moivre holds for negative \(n\); here \(n=-2\):

\(r^n\)\(=\)\(2^{-2}\)
\(=\)\(\dfrac{1}{4}\)
\(n\theta\)\(=\)\(-2\times\dfrac{\pi}{3}\)
\(=\)\(-\dfrac{2\pi}{3}\)
\(\left(2\operatorname{cis}\dfrac{\pi}{3}\right)^{-2}\)\(=\)\(\dfrac{1}{4}\operatorname{cis}\left(-\dfrac{2\pi}{3}\right)\)

\(\left(2\operatorname{cis}\dfrac{\pi}{3}\right)^{-2}=\dfrac{1}{4}\operatorname{cis}\left(-\dfrac{2\pi}{3}\right)\).

Example 4 — Modulus identities \(z\bar z=\lvert z\rvert^2\) and \(\lvert z^n\rvert=\lvert z\rvert^n\)
For \(z=3+4i\), find \(z\bar z\) and hence \(\lvert z^{3}\rvert\).
Solution

Use \(z\bar z=a^2+b^2\), which equals \(\lvert z\rvert^2\):

\(z\bar z\)\(=\)\(3^2+4^2\)
\(=\)\(9+16\)
\(=\)\(25\)
\(\lvert z\rvert\)\(=\)\(\sqrt{25}=5\)

Now apply \(\lvert z^n\rvert=\lvert z\rvert^n\) with \(n=3\):

\(\lvert z^{3}\rvert\)\(=\)\(\lvert z\rvert^{3}\)
\(=\)\(5^{3}\)
\(=\)\(125\)

\(z\bar z=25\), so \(\lvert z\rvert=5\) and \(\lvert z^{3}\rvert=125\).

Common pitfalls

Powering the argument instead of multiplying it. De Moivre gives \(\operatorname{cis}(n\theta)\), not \(\operatorname{cis}(\theta^n)\). Multiply the argument by \(n\); do not raise it to the power.
Forgetting to reduce to the principal argument. After \(n\theta\), check that the angle lies in \((-\pi,\pi]\); if not, add or subtract \(2\pi\). For example \(\dfrac{3\pi}{2}\) becomes \(-\dfrac{\pi}{2}\).
Applying De Moivre before converting to polar form. The theorem needs \(r\operatorname{cis}\theta\). Convert \(a+bi\) to polar first — you cannot power a Cartesian form directly.
Writing \(r\) instead of \(r^n\) for the modulus. The modulus of \(z^n\) is \(\lvert z\rvert^n\); for \((1+i)^6\) it is \((\sqrt{2})^6=8\), not \(\sqrt{2}\).

Frequently asked questions

What is De Moivre’s theorem?

For a complex number in polar form \(z=r\operatorname{cis}\theta\) and any integer \(n\), \((r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)\): raise the modulus to the power and multiply the argument by \(n\).

Does De Moivre’s theorem work for negative powers?

Yes. For any integer \(n\), including negative ones, \((r\operatorname{cis}\theta)^n=r^n\operatorname{cis}(n\theta)\). For example \(\left(2\operatorname{cis}\tfrac{\pi}{3}\right)^{-2}=\tfrac{1}{4}\operatorname{cis}\left(-\tfrac{2\pi}{3}\right)\).

How do you find the modulus of \(z^n\)?

Use \(\lvert z^n\rvert=\lvert z\rvert^n\): find \(\lvert z\rvert\) first, then raise it to the power. For \(z=1+i\), \(\lvert z^8\rvert=(\sqrt{2})^8=16\).

What does “principal argument” mean and why reduce to it?

The principal argument is the value of \(\arg z\) in \((-\pi,\pi]\). Multiplying by \(n\) can leave this range, so you add or subtract a multiple of \(2\pi\) to bring the answer back into it.

Why do arguments add when you multiply complex numbers?

Because \(r_1\operatorname{cis}\theta_1\times r_2\operatorname{cis}\theta_2=r_1 r_2\operatorname{cis}(\theta_1+\theta_2)\). The moduli multiply and the arguments add, which gives \(\lvert z_1 z_2\rvert=\lvert z_1\rvert\lvert z_2\rvert\) and \(\arg(z_1 z_2)=\arg z_1+\arg z_2\).

What is the quickest way to raise \(a+bi\) to a high power?

Convert to polar form, apply De Moivre’s theorem, reduce to the principal argument, then convert back to \(a+bi\). This avoids expanding a large binomial such as \((1+i)^6\).

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