Complex numbers are numbers of the form a + ib, where a and b are real numbers and i (iota) is the imaginary component and represents √(-1), and they are frequently represented in their rectangular or standard form. 2 + 3i, for example, is a complex number in which 2 is the real component and 3i is the imaginary part. Depending on the values of a and b, they can be either entirely real or purely fictitious. If a = 0 in a + ib, ib is a completely imaginary number, and if b = 0, we get a purely real number.

Polar form of Complex numbers

The modulus of a complex integer is the non-negative square root of the sum of squares of the real and imaginary parts. Modulus is written as mod(z), |z|, or |x + iy|, and is defined as follows for the complex number z = a + ib:

To represent a complex number, the polar coordinates of the real and imaginary components are written here. The symbol θ represents the angle at which the number line is inclined to the real axis, i.e. the x-axis. The length indicated by the line is known as its modulus, and it is represented by the letter r in the alphabet. The real and imaginary components are represented by a and b, respectively, while the modulus is represented by OP = r in the diagram below.

The polar form of a complex number of the type z = a + ib is represented as follows:

z = r[cos θ + i sin θ]

Here, r =   and θ = .

De Moivre Formula

To expand a complex number to its required exponent, it must first be converted to its polar form, which includes the modulus and argument as components. According to De Moivre’s Formula, for all real values of a number, say x,

(cos x + i sinx)ⁿ = cos(nx) + i sin(nx)

where, n is an integer

The complex number must be converted into its polar form (if not) to use the given formula.

Derivation

The following is how DeMoivre’s Theorem can be deduced/proven via mathematical induction:

To prove, (cos x + i sinx)ⁿ = cos(nx) + i sin(nx)     …. (1)

For n = 1, we have,

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

Suppose the formula exists for an integer, say n = k. Then we have,

        (cos x + i sin x)^k = cos(kx) + i sin(kx)      …. (2)

Now all we have to do is show that the formula holds 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 equation (2)]

        = 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}

This proves the DeMoivre’s Theorem.

Solve the following x⁴ = 5 – 5i

Solution:

It is given that, x⁴ = 5 – 5i.

The value of x is clearly, x = (5 – 5i)^1/4

Convert the complex number into its polar form.

Here, r = , θ = -π/4

The polar form of  (5 – 5i) = 

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

Thus, (5 – 5i)^1/4 = 

= 

= 

= 1.6 − 0.32i

Hence, the value of x is 1.6 − 0.32i.

Similar Problems

Question 1. Solve for x: x^(1/10) = 1 – i

Solution:

It is given that, x^(1/10) = 1 – i

The value of x is clearly, x = (1 – i)¹⁰

Convert the complex number into its polar form.

Here, r = , θ = π/4

The polar form of  (1 – i) = 

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

Thus, (1 – i)¹⁰ = 

= 

= 32 [0 + i (-1)]

= 32 (-i)

Hence, the value of x is 0 − 32i.

Question 2. Solve for x: x^(1/31) = -√3 + 3i

Solution:

It is given that, x^(1/31) = -√3 + 3i

The value of x is clearly, x = (-√3 + 3i)³¹

Convert the complex number into its polar form.

Here, r = , θ = 2π/3

The polar form of  (-√3 + 3i) = 

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

Thus, (-√3 + 3i)³¹= 

= 

= 

Hence, the value of x is .

Question 3. Solve for x: x^(1/18) = 1 + i.

Solution:

It is given that, x^(1/18) = 1 + i

The value of x is clearly, x = (1 + i)¹⁸

Convert the complex number into its polar form.

Here, r = , θ = π/4

The polar form of (1+i) = 

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

Thus, (1 + i)¹⁸ = 

= 

= 512i

Hence, the value of x is 512i.

Question 4. Solve for x: x^(1/6) = 1 + √3i.

Solution:

It is given that, x^(1/6) = 1 + √3i

The value of x is clearly, x = (1 + √3i)⁶

Convert the complex number into its polar form.

Here,  = 2, θ = π/3

The polar form of (1+i) = 

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

Thus, (1 + √3i)⁶ = 

= 

= 64(1 + 0)

= 64

Hence, the value of x is 64.

Question 5. Solve for x: x^(1/5) = 1 + i.

Solution:

It is given that, x^(1/5) = 1 + i

The value of x is clearly, x = (1 + i)⁵

Convert the complex number into its polar form.

Here, r = , θ = π/4

The polar form of (1 + i) = 

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

Thus, (1 + i)⁵ = 

= 

= -4 – 4i

Hence, the value of x is -4 – 4i.