Mathematics for Chemistry/Complex Numbers

Introduction to complex numbers

The equation:

x² + 6x + 13 = 0

Does not factorise

$x = -3 \pm \sqrt {9-13}$

$\sqrt {-4}$without complex numbers does not exist. However the number $i=\sqrt{-1}$ behaves exactly like any other number in algebra without any anomalies, allowing us to solve this problem.

The solutions are $-3\pm2i$.

2i is an imaginary number. − 3 − 2i is a complex number.

Two complex numbers (a,b) = a + ib are added by (a,b) + (c,d) = (a + c,b + d) = a + c + i(b + d).

Subtraction is obvious: (a,b) − (c,d) = (a − c,b − d).

(a,b).(c,d) = ac + i(ad + bc) − bd

Division can be worked out as an exercise. It requires(c + id)(c − id) as a common denominator. This is c² − (id)², (difference of two squares), and is c² + d².

This means

$\frac {(a,b) } {(c,d)} = \frac {ac+bd+i(cb-ad)} {c^2+d^2}$

In practice complex numbers allow one to simplify the mathematics of magnetism and angular momentum as well as completing the number system.

There is an apparent one to one correspondence between the Cartesian x - y plane and the complex numbers, x + iy. This is called an Argand diagram. The correspondence is illusory however, because say for example you raise the square root of i to a series of ascending powers. Rather than getting larger it goes round and round in circles around the origin. This is not a property of ordinary numbers and is one of the fundamental features of behaviour in the complex plane.

Plot on the same Argand diagram $2-i,~~~~3i-1,~~~~-2-2i$

Solve

x² + 4x + 29 = 0

4x² − 12x + 25 = 0

x² + 2ix + 1 = 0

(Answers -2 plus or minus 5i, 3/2 plus or minus 2i, i(-1 plus or minus root 2)

2 important equations to be familiar with, Euler's equation:

e^iθ = cos θ + isin θ

and de Moivre's theorem:

(cos θ + isin θ)ⁿ = cos nθ + isin nθ

Euler's equation is obvious from looking at the Maclaurin expansion of e^iθ.

To find the square root of i we use de Moivre's theorem.

$e^{i\frac \pi 2} = 0 +1i$

so de Moivre's theorem gives $\sqrt{e^{i\frac \pi 2}} = \cos \frac \pi {2.2} + i \sin \frac \pi {2.2}$

$= \frac 1 {\sqrt 2} + \frac 1 {\sqrt 2} i$

Check this by squaring up to give i.

The other root comes from:

$\frac {5 \pi} {4} + i \sin \frac {5 \pi} {4} = -\frac 1 {\sqrt 2} - \frac 1 {\sqrt 2} i$

de Moivre's theorem can be used to find the three cube roots of unity by

1 = cos θ + isin θ

where θ can be $0 \pm 2 \pi / n$.

Put n = 1 / 3, cos θ = − 1 / 2 and $\sin \theta = \pm \sqrt 3 / 2$

${\left( - \frac 1 2 + \frac {\sqrt 3} 2 i \right) }^3 = \left(\frac 1 4 - \frac 3 4 -\frac {\sqrt 3} 2 i \right)\left( - \frac 1 2 + \frac {\sqrt 3} 2 i \right)$

$= \left( - \frac 1 2 - \frac {\sqrt 3} 2 i \right)\left( - \frac 1 2 + \frac {\sqrt 3} 2 i \right)$

This is the difference of two squares so

${ \left( \frac 1 2 \right) }^2 - \frac 3 4 i^2~~ = ~~~ \frac 1 4 + \frac 3 4$

Similarly any collection of nth roots of 1 can be obtained this way.

Another example is to get the expressions for cos 4θ and sin 4θ without expanding cos(2θ + 2θ).

cos 4θ + isin 4θ = (cos θ + isin θ)⁴

Remember Pascal's Triangle

                          1
                      1   2   1
                    1   3   3   1
                  1   4   6   4   1
                1   5   10  10  5   1
              1   6   15  20  15  6   1

= cos ⁴θ + 4cos ³θisin θ + 6cos ²θi²sin ²θ + 4cos θi³sin ³θ + i⁴sin ⁴θ

= cos ⁴θ − 6cos ²θsin ²θ + sin ⁴θ + i(4cos ³θsin θ − 4cos θsin ³θ)

Separating the real and imaginary parts gives the two expressions. This is a lot easier than

cos(2θ + 2θ)

Use the same procedure to get

cos 6θ and sin 6θ.