Evaluating Complex Powers with De Moivre’s Theorem

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Hello! Last lesson, multiplication by a complex number meant scaling its modulus and adding its angle. A power is repeated multiplication, so it is natural to expect a pattern: the scaling accumulates, and so does the turning. De Moivre’s theorem makes that pattern precise.

Today you will use the theorem to evaluate positive, zero and negative integer powers. The aim is not just to remember a formula, but to recognise when polar form turns an intimidating expansion into a short, reliable exam solution.

From repeated multiplication to De Moivre’s theorem

Watch this brief statement of the rule before deriving it from what you already know.

Complex Numbers In Polar - De Moivre's Theorem

“Complex Numbers In Polar - De Moivre’s Theorem,” from The Organic Chemistry Tutor, states how a power acts on the modulus and argument.

Watch the power rule. Focus on the distinction between raising the modulus to a power and multiplying the angle by that power.

Write a nonzero complex number as

z = r cis θ = r (cos θ + i sin θ), r > 0.

The product rule from the previous lesson gives

z^{2} = r^{2} cis (2 θ), z^{3} = r^{3} cis (3 θ) .

Each additional factor of $z$ multiplies the modulus by $r$ and adds another $θ$ to the angle. For a positive integer $n$ , this is De Moivre’s theorem:

(r cis θ)^{n} = r^{n} cis (n θ) .

This is also a short proof by induction. The formula holds for $n = 1$ . If it holds for $n = k$ , multiplying $r^{k} cis (k θ)$ by $r cis θ$ gives $r^{k + 1} cis ((k + 1) θ)$ . Thus it holds for every positive integer $n$ .

Geometrically, taking the $n$ th power multiplies the distance from the origin by $r^{n}$ and records $n$ copies of the original turn. The resulting angle may contain several complete revolutions. That is harmless: you can reduce it by multiples of $2 π$ when you want to evaluate the sine and cosine or state a principal argument.

Turn the theorem into an exam method

An HSC-style worked example shows the full route from Cartesian form to a final Cartesian answer.

Complex Numbers — Worked Solutions (HSC Maths Extension 2) | Intuition Education

Read Example 2, “Polar form and de Moivre’s theorem,” from Intuition Education. It illustrates why identifying the modulus and argument first is more efficient than expanding a sixth power.

In Example 2 under “Solution,” find the explanatory paragraph beginning “Powers are painful in Cartesian form but easy in polar form.” Read the worked explanation through the observation about the positive real axis. Track how the first-quadrant argument is established, then why the final answer is purely real.

The example follows a useful three-stage method:

Convert: find $r = ∣ z ∣$ and a correct argument $θ$ , checking the quadrant rather than relying on a tangent value alone.
Apply De Moivre: write $z^{n} = r^{n} cis (n θ)$ . Keep this line visible; it shows exactly where the power rule was used.
Finish in the requested form: reduce the angle if helpful, then evaluate the cosine and sine. If the question asks for $x + i y$ , do not stop at an unevaluated $cis$ expression.

A different quadrant tests whether you are applying the method rather than copying the example. Let

z = - 3 + i .

Its modulus is $2$ . Since it lies in the second quadrant, an argument is $5 π /6$ , not $- π /6$ . Hence

z = 2 cis (\frac{5 π}{6}) .

To find its seventh power,

z^{7} = 2^{7} cis (\frac{35 π}{6}) = 128 cis (- \frac{π}{6}) = 643 - 64 i .

The reduction from $35 π /6$ to $- π /6$ removes three complete revolutions. A sign check offers quick reassurance: the final direction is in the fourth quadrant, so its real part should be positive and its imaginary part negative.

Zero and negative integer powers

De Moivre’s theorem also handles $n = 0$ and negative integers, provided the base is nonzero. For $n = 0$ ,

z^{0} = 1 = r^{0} cis (0) .

For a negative exponent, write $n = - m$ , where $m$ is positive. Taking the reciprocal of the positive-power result reverses its angle and reciprocates its modulus:

z^{- m} = \frac{1}{r ^{m} cis ( m θ )} = r^{- m} cis (- m θ) .

So the same boxed formula works for every integer $n$ when $r > 0$ .

For example, $z = 1 - i 3$ has modulus $2$ and argument $- π /3$ . Therefore

z^{- 2} = 2^{- 2} cis ((- 2) (- \frac{π}{3})) = \frac{1}{4} cis (\frac{2 π}{3}) = - \frac{1}{8} + \frac{3}{8} i .

The negative exponent makes the modulus $1/4$ ; the two negative signs in the angle calculation produce a positive angle. You could check the answer by squaring $1 - i 3$ and taking its reciprocal.

There is one boundary case: $0^{n} = 0$ for positive integers $n$ , but zero has no argument, so polar form is unnecessary. $0^{0}$ is not defined here, and $0$ cannot be raised to a negative integer power.

When the power is hidden inside an expression

Unfamiliar questions may not present the base directly in polar form. The previous lesson’s quotient rule can often reveal a simple modulus and angle before you raise anything to a power.

Consider

(\frac{1 + i 3}{1 - i})^{12} .

Expanding the quotient and then its twelfth power would create considerable algebra. Instead, the numerator has modulus $2$ and argument $π /3$ ; the denominator has modulus $2$ and argument $- π /4$ . Thus the quotient has modulus $2$ and an argument

\frac{π}{3} - (- \frac{π}{4}) = \frac{7 π}{12} .

De Moivre’s theorem now gives

(\frac{1 + i 3}{1 - i})^{12} = (2)^{12} cis (12 \cdot \frac{7 π}{12}) = 64 cis (7 π) = - 64 .

The useful decision was made before the calculation: retain the quotient as a quotient while finding its modulus and argument. The angle $7 π$ is an odd number of half-turns, so the answer must lie on the negative real axis.

Two cautions keep this method dependable. Raise the modulus, but multiply the angle—do not multiply the modulus by $n$ or raise the angle to a power. Also, $n θ$ is an argument of $z^{n}$ , not necessarily its principal argument. Reduce it into $(- π, π]$ only if a principal argument is requested.

Key takeaways

For nonzero $z = r cis θ$ and any integer $n$ , De Moivre’s theorem gives $z^{n} = r^{n} cis (n θ)$ . Its strength is practical as well as geometric: convert carefully, apply the power rule, and finish in the form requested by the question. Check quadrants and signs before trusting a final Cartesian answer.

Next, the direction of the problem reverses: rather than finding $z^{n}$ , you will solve equations of the form $z^{n} = w$ and account for every solution on an Argand diagram.