Demoivre’s Theorem – Integral and Rational Indices

Demoivre’s Theorem – Integral and Rational Indices

For any real number q and any integer n, (cosθ + i sinθ)ⁿ = cos nθ + i sinnθ

If n is a rational number, then one of the values of (cosθ + i sinθ)ⁿ  is cos nθ + i sinnθ.


(i) it is customary to write cisq for 1 cos nθ + i sinnθ thus we may state the Demoivre’s theorem as (cisq)ⁿ = cis(nq) if n ϵ z

(ii) (cosθ + i sinθ)⁻ⁿ = cos (-n)θ + i sin(-n)θ = cos nθ – i sinnθ provided ‘n’ is an integer.

(iii) (cosθ + i sinθ) (cosθ – i sinθ) = cos²θ – i² sin²θ = cos²θ + sin²θ = 1.

∴ \(\cos \theta +i\sin \theta =\frac{1}{\cos \theta -i\sin \theta }\).

\(\cos \theta -i\sin \theta =\frac{1}{\cos \theta +i\sin \theta }\).

(iv) \({{\left( \cos \theta -i\sin \theta  \right)}^{n}}={{\left[ \frac{1}{\cos \theta +i\sin \theta } \right]}^{n}}={{\left( \cos \theta +i\sin  \right)}^{-n}}=\cos n\theta -i\sin n\theta \). Provided ‘n’ is an integer.

(v) cisθ. cisφ = cis (θ + φ) and y + 1/y = 2 cosφ etc, then

(i) xyz … + 1/ xyz … = 2 cos (θ + φ + …)

(ii) \(\frac{x}{y}+\frac{y}{x}=2\cos \left( \theta -\phi  \right)\).

(iii) \({{x}^{m}}{{y}^{n}}+\frac{1}{{{x}^{m}}{{y}^{n}}}=2\cos
\left( m\theta +n\phi  \right)\).

(iv) \(\frac{{{x}^{m}}}{{{y}^{n}}}+\frac{{{y}^{n}}}{{{x}^{m}}}=2\cos
\left( m\theta -n\phi  \right)\).

Example: If \({{x}_{n}}=\cos \left( \frac{\pi }{{{2}^{n}}} \right)+i\sin \left( \frac{\pi }{{{2}^{n}}} \right)\), prove that x₁, x₂, x₃, … to infinity = – 1.

Solution: x₁, x₂. x₃, … to infinity

\(=\left[ \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right]\left[ \cos \frac{\pi }{{{2}^{2}}}+i\sin \frac{\pi }{{{2}^{2}}} \right]\left[ \cos \frac{\pi }{{{2}^{3}}}+i\sin \frac{\pi }{{{2}^{3}}} \right]+…….\infty \).

\(=\cos \left[ \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+\frac{\pi }{{{2}^{3}}}+… \right]+i\sin \left[ \frac{\pi }{2}+\frac{\pi }{{{2}^{2}}}+……. \right]\).

\(=\cos \left[ \frac{\pi /2}{1-1/2} \right]+i\sin \left[ \frac{\pi /2}{1-1/2} \right]\).

= cosπ + isinπ

= – 1

Roots of a complex number:

Let z = a + ib be a complex number and let r(cosθ + i sinθ) be the polar from of Z. then by Demoivre’s theorem \({{r}^{1/n}}\left[ \cos \left( \frac{\theta }{n} \right)+i\sin \left( \frac{\theta }{n} \right) \right]\) is one of the values of Z


Step 1: Write the given complex number in polar form.

Step 2: Add  to the argument

Step 3: Apply Demoivre’s theorem

Step 4: Put m = 0, 1, 2, … (n – 1) i.e., one less than the number in the denominator of the given index in the lowest form.

Example: Find ∛i

Solution: Let Z = i

We have, |z| = 1 and avg (z) = π/2

So, Z in polar form is \(\left[ \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right]\).


\({{z}^{\frac{1}{3}}}={{\left[ \cos \frac{\pi }{2}+i\sin \frac{\pi }{2} \right]}^{\frac{1}{3}}}\).

\(={{\left[ \cos \left( 2m\pi +\frac{\pi }{2} \right)+i\sin \left( 2m\pi +\frac{\pi }{2} \right) \right]}^{\frac{1}{3}}}\).

\(=\cos \left( 4m+1 \right)\frac{\pi }{6}+i\sin \left( 4m+1 \right)\frac{\pi }{6},m=0,1,2,..\).

For m = 0, we have

\({{z}^{\frac{1}{3}}}=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}=\frac{1}{2}\left( \sqrt{3}+i \right)\).

For m = 1, we have

\({{z}^{\frac{1}{3}}}=\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\).

\(=\cos \left( \pi -\frac{\pi }{6} \right)+i\sin \left( \pi -\frac{\pi }{6} \right)\).

\(=\left( -\cos \frac{\pi }{6}+i\sin \frac{\pi }{6} \right)\).

= ½ (√3 – i)

For m = 2, we have

\({{z}^{\frac{1}{3}}}=\cos \frac{9\pi }{6}+i\sin \frac{9\pi }{6}=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{6}=-i\),

Thus, the values of ∛i are \(\frac{1}{2}\left( \sqrt{3}+i \right),-\frac{1}{2}\left( \sqrt{3}-i \right)\) and – i.