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Precalculus Honors · Study guide

3.2 · Complex numbers, polar form, powers, and roots

Use modulus and argument to organize complex multiplication, powers, and all roots without losing solutions.

Precalculus Honors · Reading: Demana 10th ed. §P.6, §6.6; Stewart 8th ed. §§1.6, 8.3

Learning objectives

  • Convert a complex number to polar form.
  • Apply De Moivre’s Theorem to powers.
  • List all nth roots with distinct arguments.

The complex plane separates real and imaginary parts

The number z=a+bi\displaystyle z=a+bi corresponds to point (a,b)\displaystyle (a,b), with modulus r=a2+b2\displaystyle r=\sqrt{a^2+b^2}. Choose an argument in the correct quadrant, then write z=r(cos⁡θ+isin⁡θ)\displaystyle z=r(\cos\theta+i\sin\theta).

Arguments differing by 2π\displaystyle 2\pi describe the same nonzero number. The argument of zero is not uniquely defined.

Multiply moduli and add arguments

Polar multiplication follows the angle-addition formulas: multiply radii and add angles. Division divides moduli and subtracts arguments when the divisor is nonzero.

De Moivre’s Theorem gives [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)\displaystyle [r(\cos\theta+i\sin\theta)]^n=r^n(\cos n\theta+i\sin n\theta) for integers n\displaystyle n. Convert back to rectangular form when asked.

zn=rn(cos⁡nθ+isin⁡nθ)z^n=r^n(\cos n\theta+i\sin n\theta)

Roots distribute evenly around a circle

The nth roots of a nonzero number of modulus r\displaystyle r and argument θ\displaystyle \theta have modulus r1/n\displaystyle r^{1/n} and arguments (θ+2πk)/n\displaystyle (\theta+2\pi k)/n for k=0,…,n−1\displaystyle k=0,\ldots,n-1.

Listing only the principal argument gives one root instead of all of them. The roots are separated by angle 2π/n\displaystyle 2\pi/n, providing a geometric check on the list.

Worked examples

A polar conversion

Write 1+i3\displaystyle 1+i\sqrt{3} in polar form.

  1. The modulus is 1+3=2\displaystyle \sqrt{1+3}=2.
  2. The point lies in Quadrant I with angle π/3\displaystyle \pi/3.
  3. Thus z=2(cos⁡(π/3)+isin⁡(π/3))\displaystyle z=2(\cos(\pi/3)+i\sin(\pi/3)).

A power

Evaluate (1+i3)3\displaystyle (1+i\sqrt{3})^3.

  1. Use polar modulus 2\displaystyle 2 and angle π/3\displaystyle \pi/3.
  2. The cube has modulus 8\displaystyle 8 and angle π\displaystyle \pi.
  3. The rectangular value is −8\displaystyle -8.

Common mistakes

  • Using a first-quadrant angle for a point in another quadrant.
  • Multiplying arguments when multiplying complex numbers.
  • Listing only one root when all nth roots are requested.

Try it yourself

No calculator required.

1. Find all three complex cube roots of 8\displaystyle 8.

Three points of modulus 2 are equally spaced around the complex plane.
Compare arguments spaced by 2 pi over 3.
Show worked solution

The root modulus is 2\displaystyle 2 and arguments are 0,2π/3,4π/3\displaystyle 0,2\pi/3,4\pi/3. The roots are 2\displaystyle 2, −1+i3\displaystyle -1+i\sqrt{3}, and −1−i3\displaystyle -1-i\sqrt{3}.

Three points of modulus 2 are equally spaced around the complex plane. The roots are 2, minus 1 plus i square root of 3, and minus 1 minus i square root of 3.
The roots are 2, minus 1 plus i square root of 3, and minus 1 minus i square root of 3.

2. Find the modulus and an argument of −1+i3\displaystyle -1+i\sqrt{3}.

Show worked solution

The modulus is 2\displaystyle 2. Quadrant II gives argument 2π/3\displaystyle 2\pi/3.

3. Evaluate (1+i3)3\displaystyle (1+i\sqrt{3})^3.

Show worked solution

The polar form has modulus 2\displaystyle 2 and angle π/3\displaystyle \pi/3. Cubing gives modulus 8\displaystyle 8 and angle π\displaystyle \pi, hence −8\displaystyle -8.

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