# Maths - Exponent of Complex Number

Here we discuss two different methods to calculate the exponent Of a complex variable

• Using an infinite series below.

Let the components of the input and output planes be:

z = x + i y and w = u + i v

In this case w = ez

so:

w = e(x + i y)

by the rules of exponents:

w = exei y

applying Euler's equation we get:

w = ex(cos(y) + i sin(y))

so the u and v components are:

u = excos(y)
v = exsin(y)

## Infinite Series

An alternative method is to calculate the exponent using the series:

e(x + i y) =
 ∞ ∑ n=0
 (x + i y)n n!

On this page we derived the following expression for powers of complex numbers:

(x + i y)n=
 n ∑ k=0
 n! (n-k)! k!
(-i)k xn-k yk

So substituting in our expression for the exponent gives:

e(x + i y) =
 ∞ ∑ n=0
 n ∑ k=0
 (-i)k xn-k yk (n-k)! k!

where:

k (-i)k k!
0 1 0
1 -i 1
2 -1 2
3 i 6
4 1 24

So (-i)k cycles round every 4 entries, each step is a rotation by 90. The even values are real and the odd values are imaginary.

## Derivation of Euler's equation

In order to express this in terms of sin and cos we need to know the series for other functions (introduced on this page).

 sin(x) x - x3/3! + x5/5! ... +(-1)rx2r+1/(2r+1)! all values of x cos(x) 1 - x2/2! + x4/4! ... +(-1)rx2r/(2r)! all values of x ln(1+x) x - x2/2! + x3/3! ... +(-1)r+1xr/(r)! -1 < x <= 1 exp(x) 1 + x1/1! + x2/2! + x3/3! ... + xr/(r)! all values of x exp(-x) 1 - x1/1! + x2/2! - x3/3! ... all values of x e 1 + 1/1! + 2/2! + 3/3! =2.718281828 sinh(x) x + x3/3! + x5/5! ... +x2r+1/(2r+1)! all values of x cosh(x) 1 + x2/2! + x4/4! ... +x2r/(2r)! all values of x

exp(i y) = 1 + (i y)1/1! + (i y)2/2! + (i y)3/3!+ (i y)4/4! + (i y)5/5! + …

So if all dimensions square to positive then:

(i y)2 = -y2 = negaitive scalar

so substituting gives:

exp(i y)= 1 + (i y) -y2/2! - i y3/3! + y4/4! + i y5/5! + …

splitting up into real and imaginary parts gives:

exp(i y) = 1 -y2/2! + y4/4! +…+ i( y - y3/3! + y5/5!! + …)

exp(i y) = cos(y) + i sin(y)