Imaginary Exponents and Euler's Equation

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Imaginary Exponents and Euler's Equation

What does it mean to have an imaginary exponent? There is only one sensible way to define this. It was discovered in 1748 by the famous Swiss mathematician Leonhard Euler, who gave the following equation:

exi = cos(x) + sin(x) i

where x is any real number. This is called "Euler's Equation." Here x is the measure of an angle in radians, i is the square root of -1, and e is the base of natural logarithms. An equivalent equation had been discovered by Roger Cotes before 1722, and a related formula by Abraham DeMoivre in 1722.

Here are two different proofs of Euler's Equation.

1. Infinite series: There are infinite series expansions for each of the functions ez, cos(z), and sin(z), found from Maclaurin's Series, as follows:

ez = 1 + z + z2/2! + z3/3! + z4/4! + z5/5! + z6/6! + ... ,
cos(z) = 1 - z2/2! + z4/4! - z6/6! + z8/8! - z10/10! + ... ,
sin(z) = z - z3/3! + z5/5! - z7/7! + z9/9! - z11/11! + ... .

All three series are valid for all real numbers. We want to accept the first one as valid for all complex numbers z. Then substitute z = xi, expand, use the facts that i2 = -1, i3 = -i, i4 = 1, and so on, and collect real and imaginary parts. You'll see that the real part of exi is just the series for cos(x), and the imaginary part is just the series for sin(x).

z = cos(x) + sin(x) i

and notice that when x = 0, z = 1. Then differentiate,

dz/dx = -sin(x) + cos(x) i
dz/dx = sin(x) i2 + cos(x) i
dz/dx = [cos(x) + sin(x) i]i
dz/dx = zi
(1/z)dz/dx = i
ln(z) = xi + C

for some constant C, by indefinite integration. Now use the fact that when x = 0, z = 1, to conclude that C = 0. Thus

ln(z) = xi
z = exi
exi = cos(x) + sin(x) i

#### A consequence

A consequence of Euler's Equation is that

e i = -1
e i + 1 = 0

This remarkable equation involves the five most important constants in all of mathematics: 0, 1, i, pi (), and e. The proof is to substitute x = into Euler's Equation above.

#### An application

It is often useful to write a complex number as an exponential. This is always possible in the following way. Suppose z = a + bi, and z is not zero. Then

|z| = sqrt(a2+b2)

z/|z| is then a complex number whose absolute value is 1. Then there is some t such that

cos(t) = a/sqrt(a2+b2)
sin(t) = b/sqrt(a2+b2)
tan(t) = b/a,
t = arctan(b/a)

You can always choose t in the range 0 <= t < 2 to satisfy these conditions. There are two t values in this range with tangent b/a, which differ by . Pick t > if and only if b < 0. Pick t = if and only if b = 0 and a < 0. Then

z = sqrt(a2+b2)[a/sqrt(a2+b2) + bi/sqrt(a2+b2)]
= sqrt(a2+b2)[cos(t) + sin(t) i]
= |z|eti
= eln|z|+ti

NOTE: The value of t is not unique. You can add any integer multiple of 2 to the above value, and get another that works just as well. This has the unexpected consequence that when we evaluate complex numbers raised to complex powers, such as ii, the result is not a single complex number, but an infinite set of them.

- Robert L. Ward, for the Math Forum