Maths - Exponent of Complex Number

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

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

We start with the series:

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)


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