Complex Numbers

As we have seen, a second order constant coefficient linear homogeneous equation $y'' + p y' + q y = 0$ has solutions $y = \exp(r x)$, where $r$ is a root of the quadratic equation $r^2 + p r + q = 0$. Some quadratic equations, such as $r^2 + 1 = 0$, don't have any real roots. However, they all have complex roots. In order to study these equations, we will have to consider complex numbers.

A complex number is an expression of the form $a + b i$, where $a$ and $b$ are real numbers and $i$ is a special symbol (Engineers usually use the letter $j$ instead of $i$). While real numbers can be represented as points on a line, complex numbers can be represented as points on a plane: $a + b i$ corresponds to the point $(a,b)$. A real number $x$ can be consider as the complex number $x + 0i$ (which we would write as $x$). Thus in the complex plane, the ``$x$ axis'' consists of real numbers: we call it the real axis. The ``$y$ axis'', consisting of the numbers $y i = 0 + y i$, is called the imaginary axis, and the numbers on it are imaginary numbers. The real part of $a + b i$ is $a$, and the imaginary part is $b$. Note: in this page $a$ and $b$ are always assumed to be real numbers. We write ${\rm Re}(a+bi) = a$ and ${\rm Im}(a+bi) = b$.

Arithmetic of complex numbers follows the usual rules, with the special property $i^2 =-1$. Thus for addition and subtraction,

$$(a + b i) + (c + d i) = (a + c) + (b + d) i$$

$$(a + b i) - (c + d i) = (a - c) + (b - d) i$$

For multiplication,

$$(a + b i)(c + d i) = a c + a d i + b c i + b d i^2 = (a c - b d) + (a d + b c) i$$

e.g. $(1 + 2 i)(3 + 4 i) = (1\times 3-2\times4) + (1\times 4+2\times3)i = -5 + 10i$.

Division is a bit more complicated, and requires introducing another operation: the complex conjugate of a complex number $z = a + b i$ is $\overline{z} = a - b i$. Note that $z \overline{z} = a^2 + b^2$ is real. So to divide $z = a + b i$ by $w = c + d i$, we multiply numerator and denominator by $\overline{w}$:

$$\frac{a+bi}{c+di} = \frac{a+bi}{c+di}\frac{c-di}{c-di} = \frac{(a+bi)(c-di)}{c^2 + d^2}$$

e.g. $$\frac{1+2i}{3+4i} = \frac{(1+2i)(3-4i)}{3^2+4^2} = \frac{11 + 2 i}{25} = \frac{11}{25} + \frac{2}{25} i$$.

Complex conjugation has some important properties. To take the complex conjugate of any algebraic expression, interchange all complex variables with their complex conjugates, change every $i$ to $-i$, and leave real numbers unchanged. Thus the complex conjugate of $(z + 3 i)(\overline{z} + 2)$ is

$$\overline{(z + 3 i)(\overline{z}+ 2)} = (\overline{z} - 3 i)(z + 2)$$

Given any true equation, you can take the complex conjugate of both sides and get another true equation.

We can't compare complex numbers with ``$<$'' and ``$>$'': these only apply to real numbers. On the other hand, ``='' does apply to complex numbers: $a + b i = c + d i$ if both $a = c$ and $b = d$.

The absolute value of a complex number $z = a + b i$ is $$\vert z\vert = \sqrt{a^2 + b^2} = \sqrt{z \overline{z}}$$. This is the distance in the complex plane from $z$ to $0$.

The roots of a quadratic equation $a z^2 + b z + c = 0$ (with $a$, $b$ and $c$ real and $a \ne 0$) are

$$\frac{-b + \sqrt{b^2 - 4 ac}}{2 a} \mbox{ and } \frac{-b - \sqrt{b^2 - 4 ac}}{2 a}$$

If $b^2 - 4 a c < 0$ we interpret $$\sqrt{b^2 - 4 ac}$$ as $$\sqrt{4 ac - b^2} i$$, so we have the two complex roots

$$\frac{-b + \sqrt{4 a c - b^2} i}{2 a} \mbox{ and } \frac{-b - \sqrt{4 a c - b^2} i}{2 a}$$

Note that these roots are complex conjugates of each other.

e.g. the roots of $2 z^2 + 2 z + 5 = 0$ are $$(-2 \pm \sqrt{-36})/4 = (-1 \pm 3i)/2$$.

A much more subtle fact is that every non-constant polynomial has roots in the complex numbers. This is the Fundamental Theorem of Algebra.

We are also going to need to apply the exponential function to complex numbers. The basic formulas are $\exp(a + b i) = \exp(a) \exp(b i)$ (as you might expect) and

$$\exp(b i) = \cos b + i \sin b$$

(which may be a surprise). This is a definition, so it doesn't really need to be justified, but I might note that with this definition $\exp$ has its usual properties, of which the most important for us are $\exp(z+w) =\exp(z)\exp(w)$ and $$\frac{d}{dt} \exp(z t) = z \exp(z t)$$. If $z = a + b i$ and $w = c + d i$,

$$\exp(z+w) = \exp((a+c) + (b+d)i) = \exp(a+c) (\cos(b+d) + i \sin(b+d))$$

$$\exp(z) \exp(w) = \exp(a)(\cos b + i \sin b)\exp(c)(\cos d + i \sin d)$$

$$= \exp(a+c)(\cos b \cos d - \sin b \sin d + (\cos b \sin d + \sin b \cos d) i )$$

$$= \exp(a+c) (\cos(b+d) + i \sin(b+d)$$

and

$$\frac{d}{dt} \exp(z t) = \frac{d}{dt} (\exp(a t)(\cos bt + i\sin bt))$$

$$= (\exp(at))'(\cos bt + i \sin bt) + \exp(at) (\cos bt + i\sin bt)'$$

$$= a \exp(a t)(\cos bt + i \sin bt) + \exp(a t)(-b \sin b t + i \cos b t)$$

$$= (a + i b) \exp(a t)(\cos b t + i \sin b t) = z \exp(z t)$$

Having expressed $\exp(i b)$ in terms of $\sin b$ and $\cos b$, we can turn this around to express $\sin b$ and $\cos b$ in terms of complex exponentials:

$$\cos b = \frac{\exp(i b) + \exp(-i b)}{2} \mbox{ and } \sin b = \frac{\exp(i b) - \exp(-i b)}{2 i}$$

Note also that the trigonometric identities involving $\sin$ and $\cos$ are easy to obtain from complex exponentials. For example,

$$\cos(2 b) + i \sin(2 b) = \exp(2 b i) = (\exp(b i))^2 = (\cos b + i \sin b)^2$$

$$= \cos^2 b + 2 i \sin b \cos b - \sin^2 b$$

So $\cos(2 b) = \cos^2 b - \sin^2 b$ and $\sin(2 b) = 2 \sin b \cos b$.