Power of a Complex Number

To raise a complex number in algebraic form, \( z = a + bi \), to a power, we apply the binomial expansion formula: $$ z^n = (a+bi)^n $$ keeping in mind that the square of the imaginary unit is always equal to -1: $$ i^2 = -1 $$

In many cases, though, it's more convenient to compute the power of a complex number using its trigonometric or exponential form.

A Practical Example

Let's find the square of the complex number \( z = 1 + 3i \).

$$ z = 1 + 3i $$

We treat this as a squared binomial, \( (a + b)^2 \):

$$ z^2 = (1+3i)^2 $$

Applying the binomial square formula, \( (a + b)^2 = a^2 + 2ab + b^2 \):

$$ z^2 = 1^2 + 2(1)(3i) + (3i)^2 $$

$$ z^2 = 1 + 6i + 9i^2 $$

Since \( i^2 = -1 \):

$$ z^2 = 1 + 6i + 9(-1) $$

$$ z^2 = 1 + 6i - 9 $$

$$ z^2 = -8 + 6i $$

So, the square of \( z \) is \( z^2 = -8 + 6i \).

Example 2

Now, let's compute the cube of the complex number \( z = 1 + 3i \).

$$ z = 1 + 3i $$

We treat this as a binomial raised to the third power, \( (a + b)^3 \):

$$ z^3 = (1+3i)^3 $$

Using the binomial expansion formula, \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \):

$$ z^3 = 1^3 + 3(1^2)(3i) + 3(1)(3i)^2 + (3i)^3 $$

$$ z^3 = 1 + 9i + 3(1)(9i^2) + 27i^3 $$

$$ z^3 = 1 + 9i + 27i^2 + 27i^3 $$

Since \( i^2 = -1 \):

$$ z^3 = 1 + 9i + 27(-1) + 27i^3 $$

$$ z^3 = 1 + 9i - 27 + 27i^3 $$

$$ z^3 = -26 + 9i + 27i^3 $$

Using the fact that \( i^3 = i^2 \cdot i \):

$$ z^3 = -26 + 9i + 27(-1 \cdot i) $$

Since \( i^2 = -1 \):

$$ z^3 = -26 + 9i - 27i $$

$$ z^3 = -26 - 18i $$

So, the cube of \( z \) is \( z^3 = -26 - 18i \).

Exponentiating a Complex Number in Trigonometric Form

Given a complex number in trigonometric form: $$ z = d \cdot ( \cos \alpha + i \cdot \sin \alpha ) $$ raising it to the power of \( n \) results in another complex number where the modulus is raised to the \( n \)th power and the argument is multiplied by \( n \): $$ z^n = d^n \cdot [ \cos (n \cdot \alpha) + i \cdot \sin (n \cdot \alpha) ] $$

This result follows from De Moivre's Theorem, which establishes a fundamental link between complex numbers and trigonometric functions:

$$ ( \cos \alpha + i \sin \alpha )^n = \cos (n \alpha) + i \sin (n \alpha) $$

For exponents greater than 3, it's often easier to first convert the complex number from algebraic form to trigonometric or exponential form.

This significantly streamlines the calculation.

Example: Squaring a Complex Number

Consider the complex number:

$$ z = 1 + 3i $$

Its square is simply:

$$ z^2 = (1 + 3i) \cdot (1 + 3i) $$

To simplify the computation, we first express \( z \) in trigonometric form:

$$ z = \sqrt{1^2+3^2} \cdot \Big[ \cos \Big( \arctan \frac{3}{1} \Big) + i \sin \Big( \arctan \frac{3}{1} \Big) \Big] $$

$$ z = \sqrt{10} \cdot \Big( \cos 71.57^\circ + i \sin 71.57^\circ \Big) $$

Applying De Moivre’s Theorem with \( n = 2 \):

$$ z^2 = d^2 \cdot \Big[ \cos (2 \cdot 71.57^\circ) + i \sin (2 \cdot 71.57^\circ) \Big] $$

Substituting values:

$$ z^2 = ( \sqrt{10} )^2 \cdot \Big[ \cos (143.14^\circ) + i \sin (143.14^\circ) \Big] $$

This gives the squared complex number in trigonometric form:

$$ 10 \cdot \Big[ \cos (143.14^\circ) + i \sin (143.14^\circ) \Big] $$

Note: If needed, we can convert the result back to algebraic form using standard conversion formulas: $$ a = d \cdot \cos \alpha = 10 \cdot \cos 143.14^\circ = -8 $$ $$ b = d \cdot \sin \alpha = 10 \cdot \sin 143.14^\circ = 6 $$ Thus, in algebraic form, the squared complex number is: $$ 10 \cdot \Big[ \cos (143.14^\circ) + i \sin (143.14^\circ) \Big] = -8 + 6i $$

Proof of De Moivre’s Theorem

For a given complex number:

$$ z = d \cdot ( \cos \alpha + i \cdot \sin \alpha ) $$

Squaring both sides:

$$ z^2 = z \cdot z $$

Expanding:

$$ z^2 = \Big[ d \cdot ( \cos \alpha + i \cdot \sin \alpha ) \Big] \cdot \Big[ d \cdot ( \cos \alpha + i \cdot \sin \alpha ) \Big] $$

Using the trigonometric multiplication rule for complex numbers,

we multiply the moduli and add the arguments:

$$ d \cdot d \cdot \Big[ \cos (\alpha+\alpha) + i \sin (\alpha+\alpha) \Big] $$

$$ d^2 \cdot \Big[ \cos (2\alpha) + i \sin(2\alpha) \Big] $$

Extending this to an exponent \( n \), we multiply the complex number by itself \( n \) times:

$$ z^n = \underbrace{z \cdot z \cdots \cdot z}_{n \text{ times}} $$

Since multiplication in trigonometric form follows the pattern of multiplying moduli and summing arguments:

$$ (\underbrace{d \cdot d \cdots \cdot d}_{n \text{ times}}) \cdot \Big[ \cos ( \underbrace{\alpha + \alpha + \cdots + \alpha}_{n \text{ times}} ) + i \sin( \underbrace{\alpha + \alpha + \cdots + \alpha}_{n \text{ times}} ) \Big] $$

$$ d^n \cdot \Big[ \cos ( n \cdot \alpha) + i \sin(n \cdot \alpha) \Big] $$

Thus, for any integer \( n \), the general formula for raising a complex number to a power is:

De Moivre’s Theorem

$$ \Big[d \cdot ( \cos \alpha + i \cdot \sin \alpha ) \Big]^n = d^n \cdot \Big[ \cos (n \cdot \alpha) + i \sin(n \cdot \alpha) \Big] $$

Note: If the exponent is a negative integer, the same exponentiation rule applies as for real numbers: a negative exponent corresponds to taking the reciprocal: $$ z^{-n} = \frac{1}{z^n} $$ In trigonometric form: $$ \Big[ d \cdot (\cos \alpha + i \cdot \sin \alpha) \Big]^{-n} = \frac{1}{d^n \cdot (\cos n \alpha + i \sin n \alpha)} = \frac{1}{d^n} \cdot (\cos n \alpha - i \sin n \alpha) $$ For a detailed explanation of the algebraic steps, refer to the proof rather than repeating the derivation here.

Raising a Complex Number to a Power in Exponential Form

Given a complex number in exponential form: $$ z = d \cdot ( \cos \alpha + i \cdot \sin \alpha ) = d \cdot e^{i \alpha} $$ its \( n \)th power is given by: $$ z^n = d^n \cdot e^{i n \alpha} $$

This provides an alternative and often more convenient way to compute the \( n \)th power of a complex number.

Example

Let's take the same complex number from the previous example:

$$ z = 1+3i $$

First, we convert it to exponential form:

$$ z = \sqrt{10} \cdot e^{i 71.57^\circ} $$

To compute its square, we apply the formula with \( n = 2 \):

$$ z^2 = d^n \cdot e^{i n \alpha} $$

Substituting \( n = 2 \):

$$ z^2 = d^2 \cdot e^{i 2\alpha} $$

Since the modulus is \( \sqrt{10} \) and the argument is \( 71.57^\circ \):

$$ z^2 = (\sqrt{10})^2 \cdot e^{i (2 \cdot 71.57^\circ)} $$

$$ z^2 = 10 \cdot e^{i 143.14^\circ} $$

This is the square of the complex number \( z^2 = (1+3i)^2 \) in exponential form.

Note: To verify the result, we convert the complex number from exponential back to trigonometric form: $$ z^2 = 10 \cdot e^{i 143.14^\circ} $$ $$ z^2 = 10 \cdot ( \cos 143.14^\circ + i \sin 143.14^\circ) $$ Then, converting it to algebraic form: $$ z^2 = 10 \cdot \cos 143.14^\circ + 10 i \sin 143.14^\circ $$ $$ z^2 = 10 \cdot (-0.8) + 10 i \cdot 0.6 $$ $$ z^2 = -8 + 6i $$ which matches the result obtained earlier.

Following the same approach, we can efficiently compute higher powers as well.