How to Use De Moivre’s Theorem to Find Powers of Complex Numbers

Video Lesson: How to Use De Moivre’s Theorem to Find Powers of Complex Numbers





What is De Moivre’s Theorem?

De Moivre’s theorem is a formula used for finding powers of complex numbers. De Moivre’s theorem states that for any complex number z = r( cos(θ) + isin(θ) ), powers of this number can be calculated using zn = rn( cos(nθ) + isin(nθ) ).

To raise a complex number to a power simply raise the modulus (r) to the same power (n) and also multiply the argument (θ) by the power (n).

de moivre's theorem for calculating powers of complex numbers

De Moivre’s theorem (pronounced ‘duh mwa vruh’) provides a general equation which can be used to raise a complex number to a higher power in a much more efficient manner than via manual algebraic calculation. This is particularly true when evaluating larger powers.

In the same manner, it is used to find nth roots of complex numbers by raising the complex number to a rational power.

De Moivre’s theorem has practical applications in various fields such as engineering, physics, and mathematics.

For any complex number, z equals r times open paren cosine theta plus i sine theta close paren:

z to the power of n equals r to the power of n times open paren cosine n theta plus i sine n theta close paren

De Moivre’s theorem for raising a complex number to a power

De Moivre’s theorem can be used with positive, negative or rational powers.

How to Raise Complex Numbers to a Power Using De Moivre’s Theorem

To raise a complex number, z = r(cosθ + isinθ) to the power of n, use the formula zn = rn(cos(nθ) + isin(nθ)).

For example, 2(cos(π/3)+isin(π/3) to the power of 5 is given by 25(cos(5π/3)+isin(5π/3)). This can be evaluated and simplified as 16-16√3.

In the example of raising z equals 2 times open paren the cosine of pi over 3 plus i the sine of pi over 3 close paren to the power of 5:

  • n = 5
  • r = 2
  • 1 lines Line 1: theta equals pi over 3
how to raise a complex number to a power

Using De Moivre’s theorem: z to the power of n equals r to the power of n times open paren cosine n theta plus i sine n theta close paren

The value can be substituted in to obtain z to the fifth power equals 2 to the fifth power times open paren the cosine of open paren 5 times pi over 3 close paren plus i the sine of open paren 5 times pi over 3 close paren close paren or z to the fifth power equals 2 to the fifth power times open paren the cosine of 5 pi over 3 plus i the sine of 5 pi over 3 close paren.

  • 2 to the fifth power equals 32
  • the cosine of 5 pi over 3 equals one half
  • the sine of 5 pi over 3 equals negative the fraction with numerator the square root of 3 and denominator 2

Therefore z to the fifth power equals 2 to the fifth power times open paren the cosine of 5 pi over 3 plus i the sine of 5 pi over 3 close paren becomes z to the fifth power equals 32 times open paren one half minus the fraction with numerator the square root of 3 and denominator 2 i close paren which simplifies to z to the fifth power equals 16 minus 16 the square root of 3 i.

To raise a complex number in the form z=a+bi to the power of n, first write the complex number in the form z = r(cos nθ + isin nθ) where r = √(a2 + b2) and θ is the argument.

Then use the formula zn = rn(cos(nθ) + isin(nθ)).

To raise a complex number in the form z=a+bi to the power of n:

  1. Calculate the modulus (r) and argument (θ).
  2. Write the complex number in the form z=r(cosθ+isinθ).
  3. Raise the complex number to the power of n using zn=rn(cos(nθ)+isin(nθ)).
  4. Simplify the expression.

For example, find the exact value of open paren the square root of 3 plus i close paren to the sixth power.

Step 1. Calculate the modulus (r) and argument (θ)

The modulus of a complex number is calculated using , r equals the square root of a. squared plus b squared, where a and b are the size of the real and imaginary parts of the complex number respectively in the form z equals a. plus b i.

Therefore in z equals the square root of 3 plus i:

  • a. equals the square root of 3
  • b equals 1

Substituting these into the modulus equation, r equals the square root of open paren the square root of 3 close paren squared plus 1 squared and therefore, r equals the square root of 3 plus 1 raised to the power.

r equals the square root of 4 and so, r equals 2.

The argument is shown in the diagram below as the angle between the complex number and the positive real axis on the complex plane.

Using trigonometry, theta equals the inverse tangent of open paren the fraction with numerator 1 and denominator the square root of 3 close paren and so, theta equals pi over 6.

finding the modulus and argument of a complex number

Step 2. Write the complex number in the form z=r(cosθ+isinθ)

Using:

  • r equals 2
  • theta equals pi over 6

z equals r times open paren cosine theta plus i sine theta close paren is written as z equals 2 times open paren the cosine of pi over 6 plus i the sine of pi over 6 close paren

Step 3. Raise the complex number to the power of n using zn=rn(cos(nθ)+isin(nθ))

To raise z equals 2 times open paren the cosine of pi over 6 plus i the sine of pi over 6 close paren to the power of 6, we use De Moivre’s theorem z to the power of n equals r to the power of n times open paren cosine n theta plus i sine n theta close paren

Where:

  • r equals 2
  • theta equals pi over 6
  • n equals 6 (n is the power we are raising the complex number to)

Therefore 1 lines Line 1: n theta equals 6 times pi over 6 equals pi.

And so, z to the power of n equals r to the power of n times open paren cosine n theta plus i sine n theta close paren becomes z to the sixth power equals 2 to the sixth power times open paren cosine pi plus i sine pi close paren

Step 4. Simplify the expression

Since:

  • 2 to the sixth power equals 64
  • cosine pi equals negative 1
  • sine pi equals 0

z to the sixth power equals 2 to the sixth power times open paren cosine pi plus i sine pi close paren becomes z to the sixth power equals 64 times open paren negative 1 plus i times 0 close paren or z to the sixth power equals 64 times open paren negative 1 plus 0 close paren.

Expanding this, z to the sixth power equals negative 64.

how to use de moivres theorem to find powers of complex numbers

Find the exact value of open paren 1 plus the square root of 3 i close paren to the fourth power.

For the complex number z equals open paren 1 plus the square root of 3 i close paren

Step 1. Calculate the modulus (r) and argument (θ)

For the complex number, z equals open paren 1 plus the square root of 3 close paren:

  • a. equals 1
  • b equals the square root of 3

The modulus is given by: r equals the square root of 1 squared plus open paren the square root of 3 close paren squared and so, r equals 2.

The argument is given by theta equals the inverse tangent of open paren b over a. close paren and so theta equals the inverse tangent of open paren the fraction with numerator the square root of 3 and denominator 1 close paren which can be evaluated to obtain theta equals pi over 3.

Step 2. Write the complex number in the form z=r(cosθ+isinθ)

Using:

  • r equals 2
  • theta equals pi over 3

z equals r times open paren cosine theta plus i sine theta close paren can be written as z equals 2 times open paren the cosine of pi over 3 plus i the sine of pi over 3 close paren.

Step 3. Raise the complex number to the power of n using zn=rn(cos(nθ)+isin(nθ))

In the case of finding open paren 1 plus the square root of 3 i close paren to the fourth power, n = 4.

Therefore n theta equals 4 times pi over 3 equals 4 pi over 3.

Therefore z to the power of n equals r to the power of n times open paren cosine n theta plus i sine n theta close paren becomes z to the fourth power equals 2 to the fourth power times open paren the cosine of open paren 4 pi over 3 close paren plus i the sine of open paren 4 pi over 3 close paren close paren.

Step 4. Simplify the expression

  • 2 to the fourth power equals 16
  • the cosine of open paren 4 pi over 3 close paren equals negative one half
  • the sine of open paren 4 pi over 3 close paren equals negative the fraction with numerator the square root of 3 and denominator 2

Therefore z to the fourth power equals 2 to the fourth power times open paren the cosine of open paren 4 pi over 3 close paren plus i the sine of open paren 4 pi over 3 close paren close paren simplifies to z to the fourth power equals 16 times open paren negative one half minus the fraction with numerator the square root of 3 and denominator 2 i close paren.

Expanding this, z to the fourth power equals negative 8 minus 8 the square root of 3 i.

example of how to calculate powers of a complex number

How to Use De Moivre’s Theorem to Raise a Complex Number to a Negative Power

De Moivre’s theorem zn = rn(cos(nθ) + isin(nθ)) can be used to raise a complex number to a negative power. In this case, the value of n will be negative.

For example, find the exact value of open paren negative 1 minus i close paren to the negative 2 power.

In the form z equals a. plus b i, the complex number z equals open paren negative 1 minus i close paren has a value of a=-1 and b=-1.

Therefore r equals the square root of a. squared plus b squared becomes r equals the square root of open paren negative 1 close paren squared plus open paren negative 1 close paren squared and so, r equals the square root of 1 plus 1 and r equals the square root of 2.

The complex number is in the third quadrant of the complex plane and so, the argument is negative. It is measured clockwise from the positive real axis with an angle of theta equals negative 3 pi over 4.

The value of n is equal to the value of the power that the complex number in the question has been raised to. That is, n = -2.

Therefore z to the power of n equals r to the power of n times open paren cosine n theta plus i sine n theta close paren becomes z to the negative 2 power equals open paren the square root of 2 close paren to the negative 2 power times open paren the cosine of open paren negative 2 times negative 3 pi over 4 close paren plus i the sine of open paren negative 2 times negative 3 pi over 4 close paren close paren or z to the negative 2 power equals the fraction with numerator 1 and denominator open paren the square root of 2 close paren squared times open paren the cosine of 6 pi over 4 plus i the sine of 6 pi over 4 close paren.

Since:

  • the cosine of open paren 6 pi over 4 close paren equals 0
  • the sine of open paren 6 pi over 4 close paren equals negative 1

z to the negative 2 power equals one half times open paren 0 minus i close paren and so, z to the negative 2 power equals negative one half i.

how to use de moivre's theorem to find negative powers of a complex number

How to Find the Roots of a Complex Number

The nth roots of a complex number z=r(cosθ+isinθ) are calculated using the formula n√z = (n√r) ( cos(θ/n + 2πk/n) + i sincos(θ/n + 2πk/n) ).

De Moivre’s theorem is used with a rational power of 1 over n so that the formula for the nth root of a complex number is:

the n-th root of z equals the n-th root of r times open paren the cosine of open paren theta over n plus 2 pi k over n close paren plus i the sine of open paren theta over n plus 2 pi k over n close paren close paren

Where:

  • n is the nth root required
  • r is the modulus of the complex number
  • θ is the argument of the complex number
  • k takes values from 0 to (n-1)
how to find roots of complex numbers

For example, calculate the cube roots of z equals 8 times open paren the cosine of pi over 6 plus i the sine of pi over 6 close paren.

In this example:

  • n = 3 (since we are finding the cube root)
  • r = 8
  • theta equals pi over 6
  • k = 0, 1 and 2 (since n = 3 and we use values of k up to n-1)

Substituting these values into the n-th root of z equals the n-th root of r times open paren the cosine of open paren theta over n plus 2 pi k over n close paren plus i the sine of open paren theta over n plus 2 pi k over n close paren close paren we obtain:

the cube root of z equals the cube root of 8 times open paren the cosine of open paren the fraction with numerator open paren pi over 6 close paren and denominator 3 plus 2 pi k over 3 close paren plus i the sine of open paren the fraction with numerator open paren pi over 6 close paren and denominator 3 plus 2 pi k over 3 close paren close paren

This simplifies to:

the cube root of z equals the cube root of 8 times open paren the cosine of open paren pi over 18 plus 2 pi k over 3 close paren plus i the sine of open paren pi over 18 plus 2 pi k over 3 close paren close paren or the cube root of z equals 2 times open paren the cosine of open paren pi over 18 plus 2 pi k over 3 close paren plus i the sine of open paren pi over 18 plus 2 pi k over 3 close paren close paren.

We now substitute in values of k=0, k=1 and k=2 to obtain the three cube roots.

For k=0: we obtain

the cube root of z equals 2 times open paren the cosine of open paren pi over 18 plus the fraction with numerator 2 pi times 0 and denominator 3 close paren plus i the sine of open paren pi over 18 plus the fraction with numerator 2 pi times 0 and denominator 3 close paren close paren and so, the cube root of z equals 2 times open paren the cosine of open paren pi over 18 close paren plus i the sine of open paren pi over 18 close paren close paren

For k=1: we obtain

the cube root of z equals 2 times open paren the cosine of open paren pi over 18 plus 2 pi over 3 close paren plus i the sine of open paren pi over 18 plus 2 pi over 3 close paren close paren.

Adding the fractions to simplify we obtain:

the cube root of z equals 2 times open paren the cosine of open paren 13 pi over 18 close paren plus i the sine of open paren 13 pi over 18 close paren close paren

For k=2: we obtain

the cube root of z equals 2 times open paren the cosine of open paren pi over 18 plus the fraction with numerator 2 pi times 2 and denominator 3 close paren plus i the sine of open paren pi over 18 plus the fraction with numerator 2 pi times 2 and denominator 3 close paren close paren or the cube root of z equals 2 times open paren the cosine of open paren pi over 18 plus 4 pi over 3 close paren plus i the sine of open paren pi over 18 plus 4 pi over 3 close paren close paren.

Adding the fractions:

the cube root of z equals 2 times open paren the cosine of open paren 25 pi over 18 close paren plus i the sine of open paren 25 pi over 18 close paren close paren

calculating the cube roots of a complex number