PreCalculus

4.5. Complex exponents

As a simple example, we can show that iⁱ is a real number. To do this, we introduce Euler's formula:

e^θi = cos(θ) + sin(θ)i

The constant e is the base of natural logarithms. Now let θ = π/2. Since

cos(π/2) = 0
sin(π/2) = 1

We get

i = e^πi/2

Then, using some algebra

iⁱ = (e^πi/2)ⁱ = e^i·i·π/2 = e^-π/2

This is equal to 0.2078796 and is a real number.

Complex number raised to a complex power

We can use Euler's formula to raise a real number to a complex power. Any positive real number r can be written as e^{ln r} - so the complex power of a real number is:

r^di = (e^{ln r})^di
    = e^{(d ln r)i}
    = cos(d ln r) + sin(d ln r)i

To raise a complex number x = a + bi to the complex power y = c + di we can now do the following:

x^y = (a + bi)^{c + di}

Using Euler's formula, let

a + bi = r·e^θi

where,

     _______
r = √a² + b²
θ = tan^-1(b/a)      assure θ is in the correct quadrant

By substitution, we get

x^y = (r·e^θi)^c(r·e^θi)^di

Expanding, we get

 = r^ce^cθir^die^iidθ
 = r^ce^-dθe^cθir^di
 = r^ce^-dθ[cos(cθ) + sin(cθ)i][cos(d ln r) + sin(d ln r)i]

 = r^ce^-dθ{[cos(cθ)·cos(d ln r) - sin(cθ)·sin(d ln r)] +
         [cos(cθ)·sin(d ln r) + sin(cθ)·cos(d ln r)]i}

 = r^ce^-dθ[cos(cθ + d ln r) + sin(cθ + d ln r)i]

To summarize, given

x = a + bi
y = c + di

Then

x^y = r^ce^-dθ[cos(cθ + d ln r) + sin(cθ + d ln r)i]

where,

     _______
r = √a² + b²
θ = tan^-1(b/a)      assure θ is in the correct quadrant