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Complex Number Power Formula

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  • Last Updated : 25 Jun, 2022

Complex numbers are those with the formula a + ib, where a and b are real numbers and I (iota) is the imaginary component and represents (-1), and are often represented in rectangle or standard form. 10 + 5i, for example, is a complex number in which 10 represents the real component and 5i represents the imaginary part. Depending on the values of a and b, they might be wholly real or purely fictitious. When a = 0 in a + ib, ib is a totally imaginary number, and when b = 0, we get a, which is a strictly real number.

Complex Number Power Formula

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, DeMoivre’s theorem is applied, which states:

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

(cos x + isinx)n = cos(nx) + isin(nx), where n is any integer.

Derivation of the Formula

DeMoivre’s Theorem can be derived/proved with the help of mathematical induction as follows:

As stated previously, (cos x + isinx)n = cos(nx) + isin(nx) ⇢ (1)

For n = 1, we obtain:

(cos x + i sin x)1 = cos(1x) + i sin(1x) = cos(x) + i sin(x), which is true.

Assuming that the formula holds true for any integer, say n = k. Then,

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

Now, we just have to prove that the formula holds true for n = k + 1.

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

= (cos (kx) + i sin (kx)) (cos x + i sin x) [Using (i)]

= cos (kx) cos x − sin(kx) sinx + i (sin(kx) cosx + cos(kx) sinx)

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

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

Hence the result is proved.

Sample Questions

Question 1: Expand (1 + i)5.

Solution:

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

The 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

(\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

= −4 − 4i

Question 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\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

= 512 (-i)

= −512i

Question 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}

(\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

= 512i

Question 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]

Question 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

Question 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

Question 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})].


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