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How to Calculate a Number Raised to a Complex Power

Updated on February 19, 2014
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TR Smith is a product designer and former teacher who uses math in her work every day.

In high school you probably learned how to multiply, add, subtract, and divide complex numbers, but what you probably didn't learn is how to perform more advanced operations with complex numbers. For example, you likely learned how to simplify and calculate (5 + 3i)^4, but not 4^(5 + 3i) or (2 - 6i)^(5 + 3i), examples of numbers raised to complex exponents.

When you raise a number to a complex power, you actually end up with infinitely many answers. You can think of it as an extension of the fact that every number has two square roots (power = 1/2), three cube roots (power = 1/3), four fourth roots (power = 1/4), etc. Complex powers can be calculated by transforming the base with the formula

a + bi = r*e^(iθ) = e^[Ln(r) + iθ]

where r is the modulus of the complex number, θ is its angle in the complex plane, and i is the imaginary unit equal to the square root of -1. The way infinitely many answers arise is because θ ≡ 2π θ, 4π ± θ, 6π ± θ, ..., 2nπ ± θ, ..., so the choice of angle is not unique. Generally, θ is restricted to the interval [0, 2π) to avoid this situation. Here are several worked out examples of complex exponents.

Converting a+bi to re^(iθ) and Vice Versa

The modulus r of an complex number is the distance of the complex number when it is drawn as a vector in the complex plane. Treating a and b as vector coordinates, we have

Equation 1: r = sqrt(a^2 + b^2)

The angle θ is the radian angle measure of the vector with respect to positive real axis. To make sure θ stays in the range [0, 2π), we have to define it piecewise for various combinations of the signs of a and b. Thus we also have

Equation 2:
θ = arctan(b/a) radians, if a > 0 and b > 0
θ = arctan(b/a) + 2π radians, if a > 0 and b < 0
θ = arctan(b/a) + π, if a < 0 and b > 0
θ = arctan(b/a) + π, if a < 0 and b < 0
θ = π/2 radians, if a = 0 and b > 0
θ = 3π/2 radians, if a = 0 and b < 0
θ = 0 radians, if a > 0 and b = 0
θ = π radians, if a < 0 and b = 0

Now we want to recover a and b given the values of r and θ. If we solve Equation 2 for b and plug this into Equation 1, we can solve for a and then for b. This gives us

Equation 3: a = r*cos(θ)
Equation 4: b = r*sin(θ)

Graphical Representation of Complex Numbers in the Plane

Classic Example: How to Compute i^i

The imaginary unit raised to the imaginary unit -- i^i -- is a classic example in calculating complex exponents. First we find the modulus and angle of the imaginary unit, which are 1 and π/2 radians respectively. Thus we have

i = 1 * e^[(π/2) * i]

If we raise this to the power of i, we get

i^i = (e^[(π/2) * i])^i
= e^[(π/2) * i * i]
= e^[(π/2) * -1]
= e^-(π/2)
≈ 0.207879576

Amazingly, we have an example of an imaginary number raised to an imaginary power that equals a real number.

How to Calculate 4^(5 + 3i)

When the base is a real number and the exponent is complex, we transform the base into a power of e using the natural logarithm function. In this example, the base is 4, so we have

4 = e^Ln(4)

The angle is 0, since 4 lies on the positive real axis, so we don't have to worry about that part. Now if we use basic properties of exponents to raise this expression to the complex power 5 + 3i, we get

Helpful Resources

Complex Exponent Calculator: Outputs three distinct answers so you can compare how a change in θ affects the final answer.

Graphing Complex Numbers: Overview of how complex numbers are represented in the complex plane.

[e^Ln(4)]^(5 + 3i)
= e^[Ln(4)*(5 + 3i)]
= e^[5Ln(4) + 3Ln(4)i]
= e^(5Ln(4)) * e^(3Ln(4)i)

This is now in the form r*e^(iθ), where r equals the real number e^(5Ln(4)) and θ equals the angle 3Ln(4) radians. To put this number into the form a + bi, we need to use the second pair of complex number formulas

a = r*cos(θ)
b = r*sin(θ)

This gives us

a = e^(5Ln(4)) * cos(3Ln(4))
= -538.289

b = e^(5Ln(4)) * sin(3Ln(4))
= -871.103

Therefore we have 4^(5 + 3i) = -538.289 + -871.103i.

Harder Example: How to Compute (2 - 6i)^(5 + 3i)

From Equations 1 and 2, the modulus and angle of 2 - 6i are

r = sqrt(2^2 + 6^2) = 2*sqrt(10)
θ = arctan(-6/2) + 2π = 2π - arctan(3)

This lets us write the base of the exponent expression as

2 - 6i = e^[Ln(2*sqrt(10)) + (2π - arctan(3)i]

If we raise this to the complex power 5 + 3i, we get

(2 - 6i)^(5 + 3i)
=( e^[Ln(2*sqrt(10)) + (2π - arctan(3))i] ) ^ (5 + 3i)
= e^[ (Ln(2*sqrt(10)) + 2πi - arctan(3)i) * (5 + 3i) ]
= e^ [ 5Ln(2*sqrt(10)) - 6π + 3arctan(3) + 10πi - 5arctan(3)i + 3Ln(2*sqrt(10))i ]
= e^[ 5Ln(2*sqrt(10)) + 3arctan(3) - 6π ] * e^[ 10πi - 5arctan(3)i + 3Ln(2*sqrt(10))i ]

In this rather hairy expression, the modulus and angle are

r = e^[ 5Ln(2*sqrt(10)) + 3arctan(3) - 6π ]
= 0.0027941706

θ = 10π - 5arctan(3) + 3Ln(2*sqrt(10))
= 30.704017 radians
≡ 5.571276 radians

Using Equations 3 and 4, the complex coordinates a and b are

a = r*cos(θ) = 0.00211551
b = r*sin(θ) = -0.00182538

Therefore, one value of (2 - 6i)^(5 + 3i) is 0.00211551 - 0.00182538i. If you replace θ with θ - 2π, you can obtain another value of (2 - 6i)^(5 + 3i), which is 324842.83 - 280292.15i. Curiously, changing the angle to an equivalent angle out of the standard range results in a large change in the final answer.

Shortcut Forms

If m is a positive real number, then m^i can be calculated directly with the formula

m^i = cos(Ln(m)) + sin(Ln(m))i

Complex numbers of the form cos(x) + sin(x)i have a modulus of 1, in other words, they lie on the unit circle of the complex plane. Therefore, for any positive value of m, one value of m^i is always on the unit circle.

If n is a negative real number, we write n = e^[Ln(|n|) + πi], which gives us

n^i = (e^[Ln(|n|) + πi]) ^ i
= e^[ (Ln(|n|) + πi) * i ]
= e^[Ln(|n|)i - π]
= (e^-π)*[ cos(Ln(|n|)) + sin(Ln(|n|))i ]


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