How to Compute Complex Exponents (a+bi)^c+di

Any complex number a + bi can be written in the equivalent forms

re^iθ or
r[cos(θ) + isin(θ)],

where r is the modulus (length) of a + bi and θ is the argument (angle) of a + bi in the complex plane, measured in radians. You can find r with the formula

r = sqrt(a²+b²)

and θ with the formula

θ = arctan(b/a) + Kπ,

where K is a non-unique integer that depends on the signs of a and b.

To simplify an exponential expression of the form (a + bi)^{c + di}, first rewrite the expression as e^{(c + di)LN(a + bi)}. Then simplify it using properties of exponents and logarithms as follows:

e^{(c + di)LN(a + bi)}
= e^{(c + di)LN(reiθ)}
= e^{(c + di)[LN(r) + iθ]}
= e^{[cLN(r) - dθ] + [dLN(r) + cθ]i}

This can be split into the product of two exponentials, one with a real number power and one with an imaginary number power:

(e^{cLN(r) - dθ})(e^{[dLN(r) + cθ]i})

The first factor is a real number equal to the modulus of the complex number and the second factor contains the new angle, dLN(r) + cθ. The final answer can be written in standard form x + iy:

e^{cLN(r) - dθ}cos(dLN(r) + cθ) + ie^{cLN(r) - dθ}sin(dLN(r) + cθ)

Non-Uniqueness of Final Answers

Because any angle θ is equivalent to θ + 2nπ for any integer n, there are infinitely many ways to write the angle of a + bi and thus infinitely many complex numbers that equal (a + bi)^{c + di}. In most practical applications one takes the unique value of θ that lies in the interval [0, 2π) or the interval (-π, π].

Special Values

i ⁱ = e^iLN(eiπ/2) = e^-π/2

(1 + i)^{1 + i} = e^{(1 + i)LN(√2eiπ/4)} = (e^{LN√2 - π/4})(e^{[LN√2 + π/4]i})

© Had2Know 2010