Complex Number Power Formula

The complex number power formula helps us raise a complex number to any integer power. If a complex number is written in polar form as:

z = r(\cos\theta + i\sin\theta) = re^{i\theta}

where r is the modulus and θ is the argument, then by De Moivre’s Theorem:

z^n = r^n(\cos(n\theta) + i\sin(n\theta)) = r^n e^{in\theta}

This means the modulus is raised to the power n, and the argument is multiplied by n, making power calculations easy.

De Moivre's Theorem

To expand a complex number according to its specified exponent, it must first be transformed to its polar form, which has the modulus and argument as components. After that, De Moivre's theorem is applied, which states:

De Moivre's Formula states that for all real values of a number, say x,

General form: (\cos x + i\sin x)^n = \cos(nx) + i\sin(nx), \quad n \in \mathbb{Z}

Exponential form: (e^{ix})^n = e^{inx}

Formula Derivation

DeMoivre's Theorem can be derived with the help of Mathematical Induction as follows:

P(n): (cos x + i sin x)ⁿ = cos(nx) + isin(nx) ⇢ (1)

For n = 1, we have

P(1) = (cos x + i sin x)¹ 

P(1) = cos(1x) + i sin(1x) 

P(1) = cos(x) + i sin(x)

That is true and thus, P(1) is true.

Assuming P(k) is true, i.e.

P(k) = (cos x + i sin x)^k = cos(kx) + i sin(kx) ⇢  (2)

Now, we just have to prove that the P(k+1) is also true.

P(k+1) = (cos x + i sin x)^k+1 

⇒ P(k+1) = (cos x + i sin x)^k (cos x + i sin x)

⇒ P(k+1) = (cos (kx) + i sin (kx)) (cos x + i sin x) [Using (i)]

⇒ P(k+1) = cos (kx) cos x − sin(kx) sinx + i (sin(kx) cosx + cos(kx) sinx)

⇒ P(k+1) = cos {(k + 1)x} + i sin {(k + 1)x}

⇒ P(k+1) = (cos x + i sin x)^k+1 = cos {(k + 1)x} + i sin {(k + 1)x}

Thus, P(k+1) is also true, thus by the principal of mathematical induction, P(n) is true.

Hence the result is proved.

Also Check

• Absolute Value of a Complex Number
• Polar Representation of Complex Numbers
• Z Bar in Complex Numbers

Solved Example on Complex Number Power Formula

Example 1: Expand (1 + i)⁵.

Solution:

Given, 

• r = √(1² + 1²) = √2
• θ = π/4

Polar form of (1 + i) =(2\sqrt{2})^{4}-(\sqrt{2})^{4}i

According to De Moivre's Theorem 

(cosθ + sinθ)ⁿ = cos(nθ) + i sin(nθ)

Thus, 

(1 + i)⁵ = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^5

⇒ (1 + i)⁵ = (\sqrt{2})^{5}[cos(\frac{5\pi}{4})+i\ sin(\frac{5\pi}{4})]\\ =(\sqrt{2})^{5}[cos(\pi+\frac{\pi}{4})+i\ sin(\pi+\frac{\pi}{4})]\\ =(\sqrt{2})^{5}[-cos(\frac{\pi}{4})-i\ sin(\frac{\pi}{4})]\\ =(\sqrt{2})^{5}[-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i]\\ =(\sqrt{2})^{4}-(\sqrt{2})^{4}i

⇒ (1 + i)⁵ = −4 − 4i

Example 2: Expand (2 + 2i)⁶.

Solution:

Here, r = \sqrt{(2^2+2^2)} = 2\sqrt{2}     , θ = π/4

The polar form of  (2 + 2i) = [2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]

According to De Moivre's Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + i sin(nθ).

Thus, (2 + 2i)⁶ = [2\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^6

⇒ (2 + 2i)⁶ = (2\sqrt{2})^{6}[cos(\frac{6\pi}{4})+i\ sin(\frac{6\pi}{4})]\\ =(2\sqrt{2})^{6}[cos(\frac{3\pi}{2})+i\ sin(\frac{3\pi}{2})]\\ =(2\sqrt{2})^{6}[cos(\pi+\frac{\pi}{2})+i\ sin(\pi+\frac{\pi}{2})]\\ =(2\sqrt{2})^{6}[cos(\frac{\pi}{2})-i\ sin(\frac{\pi}{2})]\\ =(2\sqrt{2})^{6}[0-i]\\ =-(2\sqrt{2})^{6}i

⇒ (2 + 2i)⁶ = 512 (-i) = −512i

Example 3: Expand (1 + i)¹⁸.

Solution:

Here, r = \sqrt{(1^2+1^2)} = \sqrt{2}     , θ = π/4

The polar form of (1+i) = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]

According to De Moivre's Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + isin(nθ).

Thus, (1 + i)¹⁸ = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{18}

⇒ (1 + i)¹⁸= (\sqrt{2})^{18}[cos(\frac{18\pi}{4})+i\ sin(\frac{18\pi}{4})]^{18}\\ =(\sqrt{2})^{18}[cos(\frac{9\pi}{2})+i\ sin(\frac{9\pi}{2})]\\ =(\sqrt{2})^{18}[cos(4\pi+\frac{\pi}{2})+i\ sin(4\pi+\frac{\pi}{2})]\\ =(\sqrt{2})^{18}[cos(\frac{\pi}{2})+i\ sin(\frac{\pi}{2})]\\ =(\sqrt{2})^{18}[0+i]\\ =(\sqrt{2})^{18}i

⇒ (1 + i)¹⁸ = 512i

Example 4: Expand (-√3 + 3i)³¹.

Solution:

Here, r = \sqrt{((-\sqrt{3})^2+3^2)} = 2\sqrt{3}     , θ = 2π/3

The polar form of  (-√3 + 3i) = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]

According to De Moivre's Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + i sin(nθ).

Thus, (-√3 + 3i)³¹= [2\sqrt{3}(cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4}))]^{31} = (2\sqrt{3})^{31}[cos(\frac{31\pi}{4})+i\ sin(\frac{31\pi}{4})]\\ =(2\sqrt{3})^{31}[cos(8\pi-\frac{\pi}{4})+i\ sin(8\pi-\frac{\pi}{4})]\\ =(2\sqrt{3})^{31}[cos(\frac{\pi}{4})-i\ sin(\frac{\pi}{4})]\\ =(2\sqrt{3})^{31}[\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i]

Example 5: Expand (1 - i)¹⁰.

Solution: 

r = \sqrt{(1^2+(-1)^2)} = \sqrt{2}     , θ = π/4

The polar form of  (1 - i) = \sqrt{2}[cos(\frac{\pi}{4})+i \ sin(\frac{\pi}{4})]

According to De Moivre's Theorem: (cosθ + sinθ)ⁿ = cos(nθ) + i sin(nθ).

Thus, (1 - i)¹⁰ = [\sqrt{2}cos(\frac{\pi}{4})+i\ sin(\frac{\pi}{4})]^{10}

= [\sqrt{2}(cos(\frac{π}{4})+ i sin(\frac{π}{4}))]^{10}\\ = (\sqrt2)^{10}[cos(\frac{10π}{4})+i\ sin(\frac{10π}{4})]\\ =(\sqrt2)^{10}[cos(\frac{5π}{2})+i\ sin(\frac{5π}{2})]\\ =(\sqrt2)^{10}[cos(2\pi+\frac{π}{2})+i\ sin(2\pi+\frac{π}{2})]\\ =(\sqrt2)^{10}[cos(\frac{π}{2})-i\ sin(\frac{π}{2})]\\

= 32 [0 + i(-1)]

= 32 (-i)

= -32i

Example 6: Simplify (1 + √3i)⁶.

Solution:

Modulus of (1 + √3i)⁶ = \sqrt{1^2+(\sqrt{3})^2} = 2

Argument = tan^-1(√3/1) = tan^-1(√3) = π/3

⇒ Polar form = 2[cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3})]           

Now, (1 + √3i)⁶ = [2(cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3}))]^6      

As per DeMoivre's theorem, (cos x + isinx)ⁿ = cos(nx) + isin(nx).

⇒ [2(cos(\frac{\pi}{3})+i\ sin(\frac{\pi}{3}))]^6  

= 2^6(cos(\frac{6\pi}{3})+i\ sin(\frac{6\pi}{3}))

= 64 (cos 2π + i sin 2π)

= 64(1 + 0)

= 64

Example 7: Simplify i^√3.

Solution:

Modulus = r = \sqrt{0^2+1^2}      = 1

Argument = tan^-1[1/0] = π/2

Polar Form = r[cosθ + isinθ] = 1[cos(\frac{\pi}{2}) +i\ sin(\frac{\pi}{2})]  

Now, i^{√3} = [cos(\frac{\pi}{2}) + i\ sin(\frac{\pi}{2})]^{\sqrt3}

As per DeMoivre's theorem: (cosθ + isinθ)ⁿ = cos(nθ) + isin(nθ).

⇒  [cos(\frac{\pi}{2}) + i\ sin(\frac{\pi}{2})]^{\sqrt3}  

= [cos(\frac{\sqrt3\pi}{2}) + i\ sin(\frac{\sqrt3\pi}{2})].

Practice Problems on Complex Number Power Formula

Problem 1: Find the value of (2 − 3i)⁴.

Problem 2: Calculate (3+4i)³.

Problem 3:  Compute the square root of 3 + 4i in polar form.