Uses of De Moivre’s Theorem

De Moivre’s Theorem is essential since it can easily find the powers and roots of complex numbers. This theorem helps us find the power of any complex number in the polar form. 

The primary use of De Moivre’s Theorem is to obtain the relationship between the powers of trigonometric functions (e.g.- cos4x, sin2 x) and trigonometric functions of multiple angles (e.g.- cos 7x, sin 3x). Another prominent use of De Moivre’s Theorem is to obtain the roots of the polynomial equations.

It can help you raise complex numbers to the highest powers, proving the famous trigonometric identities. This theorem can also solve any rational number—positive numbers, negative numbers, and fractions.

De Moivre’s Theorem is also known as De Moivre’s Formula and De Moivre’s Identity. It was formulated by the great mathematician Abraham de Moivre. He was very famous and made many contributions to mathematics, mainly in the theory of algebra and probability.

Importance of De Moivre’s Theorem

De Moivre’s Theorem is a fundamental theorem when solving complex numbers. Finding the roots of complex numbers without this theorem would be very difficult, since solving complex numbers in a straightforward manner is a challenging and lengthy process. 

De Moivre’s Theorem Questions

Here are some questions based on De Moivre Theorem:

• Question 1

If z = (cosθ + i sinθ ), show that zn + 1/ zn = 2 cos nθ and zn – [1/ zn] = 2i sin nθ

Solution

Let z = (i sinθ + cosθ)

By de Moivre’s theorem,

zn = (cosθ + i sinθ )n = cos nθ + i sin nθ 

• Question 2

A Proof z = (√3/2 + i/2)5 + (√3/2 – i/2)5 = 2cos(150°)

Solution

Here we have to put √3/2 = rcosθ and 1/2 = rsinθ

Now we have (rcosθ)2 + (rsinθ)2 = 3/4 + 1/4 = 1

so r2(cos2θ + sin2θ) = 1 

Hence, r2 = 1 ⇒ r =1

Now rsinθ = 1/2 ⇒ sinθ = 1/2

⇒ θ = 30°

Now z = (cosθ + isinθ)5 + (cosθ – isinθ)5

= cos5θ + isin5θ + cos5θ – isin5θ

=2cos5θ ⇒ 2cos150°

B If cos α + cosβ + cosγ = sinα +sinβ +sinγ = 0

Then prove that

cos3α + cos3β +cos3γ = 3cos(α+β+γ)

Now we take sinα + sinβ + sinγ = 0, multiply both sides with i and then add it to cosα + cosβ + cosγ

so, we have i(sinα + sinβ +sinγ) + (cosα + cosβ + cosγ) = 0

= cosα + isinα +cosβ + isinβ + cosγ + isinγ = 0

Now let z1 = cosα + isinα

z2 = cosβ + isinβ

z3 = cosγ + isinγ

And we have z1 +z2 +z3 = 0

Through algebra, we know that z13 + z23 + z33 – 3z1.z2.z3 = (z1 + z2 + z3).(z12 + z22 + z32 – z1z2 – z2z3 – z3z1)

So z13 + z23 + z33 = 3z1z2z3

Now we have (cosα + isinα)3 + (cosβ + isinβ)3 + (cosγ + isinγ)3

= 3(cosα + isinα)(cosβ + isinβ)(cosγ + isinγ)

⇒ (cos3α + isin3α) + (cos3β + isin3β) + (cos3γ + isinγ3γ)

= 3[cos(α+β +γ) + isin(α+β +γ)] from (i) {see theory part}

Now equate real part to real part we have:

cos3α + cos3β + cos3γ = 3cos(α+β +γ)

• Question 3

(cos2θ + isin2θ)-5.(cos3θ – isin3θ)6.(sinθ – icosθ)3

Solution 

Firstly sinθ – icosθ can be written as -i2sinθ – cosθ = -i(cosθ + isinθ)

= (cos(-10θ) + isin(-10θ)).(cos18θ – isin18θ).(-i)3(cos3θ + isin3θ) 

(here cos18θ = cos(-18θ))

Now from (I) (see theory) we have 

i*(Cos(-10θ -18θ+ 3θ) + isin(-10θ -18θ -+3θ))

= i( cos(-25θ) + isin(-25θ)) 

= i(cos25θ – isin25θ)

• Question 4

Compute (cos2θ – isin2θ)7.(cos3θ + isin3θ)-5/(cos4θ + isin4θ)12

Solution

(cos14θ – isin14θ).(cos(-15θ) + isin(-15θ))/(cos48θ + isin48θ)

Now from eulers form we have

= e-14iθ.e-15iθ / e48iθ

= e-77iθ

= cos77θ – isin77θ 

Conclusion

You can solve an equation and find the roots in the polar form of any complex number, even if it is a positive number, negative number, and fraction. This is possible because of Abraham de Moivre’s theorem.

This theorem is one of the central theorems to solve high-level questions.