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Pre-Calculus Complex numbers (advanced)

DeMoivre's Theorem (powers and roots of complex numbers)

20 practice questions 0 video lessons Theory + worked examples

DeMoivre's Theorem

California Pre-Calculus • Standard N-CN.5 • Complex Numbers

DeMoivre's Theorem is a topic in Complex Numbers in the California Common Core State Standards. It is aligned to Standard N-CN.5, which requires students to use DeMoivre's Theorem to find the powers and roots of complex numbers.

DeMoivre's Theorem raises a complex number to a power as \((r\,\text{cis}\,\theta)^n=r^n\,\text{cis}(n\theta)\) and finds its \(n\) equally spaced roots.

California Pre-Calculus › Complex Numbers › DeMoivre's Theorem  —  Standard N-CN.5

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Theory

DeMoivre's Theorem makes powers of complex numbers easy in polar form:
\[\big(r\,\text{cis}\,\theta\big)^n=r^n\,\text{cis}(n\theta).\]

Raise the modulus to the power and multiply the argument by \(n\).

Reversing it gives the \(n\)th roots: a nonzero complex number has exactly \(n\) of them, equally spaced by \(\dfrac{360^\circ}{n}\) around a circle of radius \(r^{1/n}\).

All \(n\) roots share the same modulus \(r^{1/n}\); only their arguments differ, by equal steps.
DeMoivre's Theorem DeMoivre's Theorem DeMoivre's Theorem zⁿ = rⁿ cis(nθ) nth roots: r^(1/n) cis((θ+360°k)/n)
DeMoivre's Theorem for powers and for \(n\)th roots.
The nth roots are equally spaced on a circle The n complex nth roots of a number lie on a circle, equally spaced by 360 over n degrees. roots spaced 360°/n apart
The \(n\)th roots sit on a circle, equally spaced.

Powers and roots in polar form:

\[\big(r\,\text{cis}\,\theta\big)^n=r^n\,\text{cis}(n\theta)\]
\[\sqrt[n]{z}=r^{1/n}\,\text{cis}\!\left(\dfrac{\theta+360^\circ k}{n}\right),\quad k=0,1,\dots,n-1\]
z to the n equals r to the n cis of n theta; the nth roots use r to the one over n and arguments spaced evenly
Use every \(k\) from 0 to \(n-1\) to get all \(n\) distinct roots.

How to use DeMoivre's Theorem

  1. Convert the number to polar form.
  2. Power: raise \(r\) to the power, multiply \(\theta\) by \(n\).
  3. Roots: take \(r^{1/n}\); use arguments \(\dfrac{\theta+360^\circ k}{n}\) for \(k=0,\dots,n-1\).
  4. Convert back to rectangular form if required.
Example 1 — A power
Compute \((2\,\text{cis}\,30^\circ)^4\).
Solution

Raise the modulus to the power and multiply the argument by it.

\(=\)\(2^4\,\text{cis}\,(4\cdot 30^\circ)\)
\(=\)\(16\,\text{cis}\,120^\circ\)
result is 16 cis 120 degrees
Example 2 — Power of a rectangular number
Compute \((1+i)^6\).
Solution

Convert to polar (\(r=\sqrt2,\ \theta=45^\circ\)), then apply DeMoivre.

\(=\)\((\sqrt2)^6\,\text{cis}\,(6\cdot 45^\circ)\)
\(=\)\(8\,\text{cis}\,270^\circ\)
\(=\)\(-8i\)
result is negative 8 i
Example 3 — Number of nth roots
How many distinct cube roots does a nonzero complex number have, and how are they arranged?
Solution

A number has exactly \(n\) distinct \(n\)th roots, equally spaced around a circle.

\(n\)\(=\)\(3\ \text{roots}\)
\(\text{spacing}\)\(=\)\(\dfrac{360^\circ}{3}=120^\circ\)
three cube roots, spaced 120 degrees apart
Example 4 — Find the roots
Find the square roots of \(9\,\text{cis}\,60^\circ\).
Solution

Take \(\sqrt9=3\) for the modulus; the arguments are \(\dfrac{60^\circ+360^\circ k}{2}\) for \(k=0,1\).

\(k=0\)\(:\)\(3\,\text{cis}\,30^\circ\)
\(k=1\)\(:\)\(3\,\text{cis}\,210^\circ\)
square roots are 3 cis 30 degrees and 3 cis 210 degrees

Common pitfalls

A number has \(n\) distinct \(n\)th roots, not one — step \(k\) through \(0\) to \(n-1\).
Only the argument gets divided for roots; the modulus is \(r^{1/n}\).
Convert to polar before applying DeMoivre. The theorem is stated in polar form.

Frequently asked questions

What is DeMoivre's Theorem?

\((r\,\text{cis}\,\theta)^n=r^n\,\text{cis}(n\theta)\): to raise a complex number to a power, raise the modulus and multiply the argument by \(n\).

How many nth roots does a complex number have?

Exactly \(n\), equally spaced by \(\dfrac{360^\circ}{n}\) around a circle of radius \(r^{1/n}\).

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

Take \(r^{1/n}\) as the modulus and arguments \(\dfrac{\theta+360^\circ k}{n}\) for \(k=0,1,\dots,n-1\).

Why is polar form needed for powers and roots?

Because DeMoivre's Theorem acts on the modulus and argument directly, which only appear in polar form.

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