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

(a+bi)c+di Calculator
a =           b =

c =           d =

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

re 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(re)
= e(c + di)[LN(r) + ]
= e[cLN(r) - dθ] + [dLN(r) + ]i

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

(ecLN(r) - )(e[dLN(r) + ]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) + . The final answer can be written in standard form x + iy:

ecLN(r) - cos(dLN(r) + ) + iecLN(r) - sin(dLN(r) + )

### 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 = eiLN(eiπ/2) = e-π/2

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