Precalculus Polar Form and De Moivre’s Theorem Lesson (Free PDF)

Precalculus math lesson: Polar Form and De Moivre's Theorem – Maths PDF

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This polar complex lesson for Precalculus covers polar form and de moivre's theorem (Common Core N-CN.B.4, N-CN.B.5, N-CN.B.6), with a free PDF to print.

The polar form of a complex number is z = r(cos theta + i sin theta), where r = |z| is the modulus and theta is the argument. Multiplying multiplies the moduli and adds the angles. De Moivre’s theorem gives powers, z^n = r^n (cos n theta + i sin n theta), and n evenly spaced nth roots.

Precalculus · Common Core State Standards: N-CN.B.4, N-CN.B.5, N-CN.B.6

Key facts in this polar complex lesson

Key facts

z = a + bi = r(cos theta + i sin theta) with r = sqrt(a^2 + b^2) and tan theta = b/a.
Product: multiply moduli, add arguments. Quotient: divide moduli, subtract arguments.
De Moivre: [r cis theta]^n = r^n cis(n theta).
nth roots: (r)^(1/n) cis((theta + 360 k)/n), k = 0, …, n – 1.
The n roots form a regular n-gon on a circle of radius r^(1/n).

What is the polar form of a complex number?

A complex number \(z=a+bi\) is the point \((a,b)\) in the complex plane. Its distance from the origin is the modulus \(r=|z|=\sqrt{a^{2}+b^{2}}\), and the angle from the positive real axis is the argument \(\theta\), with \(\tan\theta=\frac{b}{a}\) in the correct quadrant. Because \(a=r\cos\theta\) and \(b=r\sin\theta\), the polar form is \(z=r(\cos\theta+i\sin\theta)\), often written \(r\,\text{cis}\,\theta\).

\(z=1+\sqrt3\,i\) has modulus \(2\) and argument \(60^{\circ}\), so \(z=2(\cos60^{\circ}+i\sin60^{\circ})\).

How do you multiply and divide in polar form?

Product and quotient rules

If \(z_{1}=r_{1}\text{cis}\,\theta_{1}\) and \(z_{2}=r_{2}\text{cis}\,\theta_{2}\), then \(z_{1}z_{2}=r_{1}r_{2}\,\text{cis}(\theta_{1}+\theta_{2})\) and \(\frac{z_{1}}{z_{2}}=\frac{r_{1}}{r_{2}}\,\text{cis}(\theta_{1}-\theta_{2})\).

Multiplication multiplies the moduli and adds the angles. Geometrically it scales and rotates. For example, \(2\,\text{cis}\,30^{\circ}\cdot3\,\text{cis}\,45^{\circ}=6\,\text{cis}\,75^{\circ}\).

The product has modulus 2 times 3 = 6 and argument 30° + 45° = 75°.

What is De Moivre’s theorem?

Repeated multiplication gives \([r(\cos\theta+i\sin\theta)]^{n}=r^{n}(\cos n\theta+i\sin n\theta)\). To compute \((1+i)^{8}\): \(1+i=\sqrt2\,\text{cis}\,45^{\circ}\), so \((1+i)^{8}=(\sqrt2)^{8}\text{cis}\,360^{\circ}=16\).

How do you find the nth roots of a complex number?

A nonzero number \(z=r\,\text{cis}\,\theta\) has exactly \(n\) distinct \(n\)th roots: \(\sqrt[n]{r}\,\text{cis}\left(\frac{\theta+360^{\circ}k}{n}\right)\) for \(k=0,1,\text{ … },n-1\). They lie on a circle of radius \(\sqrt[n]{r}\), equally spaced \(\frac{360^{\circ}}{n}\) apart. For the cube roots of 8, use \(r=8\), \(\theta=0^{\circ}\): \(2\,\text{cis}\,0^{\circ}\), \(2\,\text{cis}\,120^{\circ}\), \(2\,\text{cis}\,240^{\circ}\).

The three cube roots of 8 form an equilateral triangle on a circle of radius 2.

The roots of unity deserve special attention. The sixth roots of 1 are \(\text{cis}\,0^{\circ}\), \(\text{cis}\,60^{\circ}\), and so on in steps of \(60^{\circ}\). They sit at the corners of a regular hexagon inscribed in the unit circle, and they add up to zero. Roots of unity appear in signal processing, in geometry (constructing regular polygons) and in solving equations of the form \(z^{n}=1\).

The sixth roots of unity form a regular hexagon.

Example (converting): To write \(-3+3i\) in polar form, find \(r=\sqrt{9+9}=3\sqrt2\). The point lies in quadrant II, and \(\arctan\frac{3}{-3}=-45^{\circ}\) would put it in quadrant IV, so add \(180^{\circ}\) to get \(\theta=135^{\circ}\). Always check the quadrant, because a calculator returns an angle between \(-90^{\circ}\) and \(90^{\circ}\) only.

The steps are: write \(z\) in polar form, take the \(n\)th root of \(r\), then divide the argument plus each multiple of \(360^{\circ}\) by \(n\). The roots of 1, called roots of unity, form a regular polygon inscribed in the unit circle.

Polar complex lesson: questions and answers

How do you convert a complex number to polar form?

Find the modulus r = sqrt(a^2 + b^2) and the argument theta with tan theta = b/a, choosing the quadrant that matches the point. For -3 + 3i, r = 3 sqrt 2 and theta = 135 degrees.

Why do angles add when you multiply?

The product rule for cosine and sine uses the angle addition formulas, so multiplying cis A by cis B gives cis(A + B). Geometrically, multiplying rotates by the other number’s argument.

What is De Moivre’s theorem used for?

It raises a complex number to a power quickly: take the power of the modulus and multiply the angle by the exponent. It also leads to the formula for nth roots.

How many nth roots does a complex number have?

Exactly n, all with the same modulus. Their arguments differ by 360 degrees divided by n.

What is cis theta?

It is shorthand for cos theta + i sin theta, which saves writing when working with polar form.

How to use this polar complex lesson

Read the definitions first, then follow the worked method line by line and redo the example on paper without looking. When every step makes sense, move on to the worksheet, then check yourself with the quiz.

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