# Mathematics for Chemistry/Complex Numbers
## Introduction to complex numbers

The equation:
$x^{2}+6x+13=0$
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 $-3\pm 2i$.
$2i$ 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 $c^{2}-(id)^{2}$, (difference of two squares), and is $c^{2}+d^{2}$.
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
$x^{2}+4x+29=0$
$4x^{2}-12x+25=0$
$x^{2}+2ix+1=0$
(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:
$e^{i\theta }=\cos \theta +i\sin \theta$
and de Moivre's theorem:
${(\cos \theta +i\sin \theta )}^{n}=\cos n\theta +i\sin n\theta$
Euler's equation is obvious from looking at the Maclaurin expansion of $e^{i\theta }$.
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
$1=\cos \theta +i\sin \theta$
where $\theta$ can be $0\pm 2\pi /n$.
Put $n=1/3$, $\cos \theta =-1/2$ 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 $n$th roots of 1 can be obtained this way.
Another example is to get the expressions for $\cos 4\theta$ and $\sin 4\theta$ without expanding $\cos(2\theta +2\theta )$.
$\cos 4\theta +i\sin 4\theta ={(\cos \theta +i\sin \theta )}^{4}$
Remember Pascal's Triangle
$=\cos ^{4}\theta +4\cos ^{3}\theta i\sin \theta +6\cos ^{2}\theta i^{2}\sin ^{2}\theta +4\cos \theta i^{3}\sin ^{3}\theta +i^{4}\sin ^{4}\theta$
$=\cos ^{4}\theta -6\cos ^{2}\theta \sin ^{2}\theta +\sin ^{4}\theta +i(4\cos ^{3}\theta \sin \theta -4\cos \theta \sin ^{3}\theta )$
Separating the real and imaginary parts gives the two expressions. This is a lot easier than
$\cos(2\theta +2\theta )$
Use the same procedure to get
$\cos 6\theta$ and $\sin 6\theta$.