A-level Mathematics/MEI/FP2/Complex Numbers

Modulus-argument form

Polar form of a complex number

It is possible to express complex numbers in polar form. The complex number z in the diagram below can be described by the length r and the angle θ of its position vector in the Argand diagram.

[Argand diagram]

The distance r is the modulus of z, $|z|\,$. The angle θ is measured from the positive real axis and is taken anticlockwise. Adding any whole multiple of 2π however, would give the same vector so a complex number's principal argument, $\arg{z}\,$, is where $- \pi < \theta \leq \pi$. The following examples demonstrate this in each quadrant.

The following Argand diagram shows the complex number $z = 1 + \sqrt{3}j$.

[Argand diagram]

$|z| = \sqrt{1^2 + \sqrt{3}^2} = \sqrt{1 + 3} = 2$

$\arg z = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}\,$

This Argand diagram shows the complex number $z = 2 - 4j\,$.

This Argand diagram shows the complex number $z = - 8 + 5j\,$.

This Argand diagram shows the complex number $z = - 5 - 6j\,$.

When we have a complex number $z=x+yj\,$ in polar form $(r, \theta)\,$ we can use $x = r\cos{\theta}\,$ and $y = r\sin{\theta}\,$ to write it in the form: $z = r(\cos{\theta}+j\sin{\theta})\,$. This is the modulus-argument form for complex numbers.

Multiplication and division

The polar form of complex numbers can provide a geometrical interpretation of the multiplication and division of complex numbers.

Multiplication

Take two complex numbers in polar form,

$\omega_1 = r_1(\cos{\theta_1}+j\sin{\theta_1})\,$

$\omega_2 = r_2(\cos{\theta_2}+j\sin{\theta_2})\,$

and then multiply them together,

$\begin{array}{rl} \omega_1\omega_2 & = r_1r_2(\cos{\theta_1}+j\sin{\theta_1})(\cos{\theta_2}+j\sin{\theta_2}) \\ & = r_1r_2((\cos{\theta_1}\cos{\theta_2}-\sin{\theta_1}\sin{\theta_2})+j(\sin{\theta_1}\cos{\theta_2}+\cos{\theta_1}\sin{\theta_2})) \\ & = r_1r_2(\cos{(\theta_1 + \theta_2)}+j\sin{(\theta_1 + \theta_2)}) \end{array} \,$

The result is a complex number with a modulus of $r_1r_2\,$ and an argument of $\theta_1+\theta_2\,$. This means that:

$|\omega_1\omega_2| = |\omega_1||\omega_2|\,$

$\arg{(\omega_1\omega_2)} = \arg{\omega_1} + \arg{\omega_2}\,$

De Moivre's theorem

Using the multiplication rules we can see that if

$z = \cos{\theta} + j\sin{\theta}\,$

then

$z^2 = \cos{2\theta} + j\sin{2\theta}\,$

$z^3 = \cos{3\theta} + j\sin{3\theta}\,$

De Moivre's theorem states that this holds true for any integer power. So,

$z^n = \cos{n\theta} + j\sin{n\theta}\,$

Complex exponents

Definition

If we let $z = \cos{\theta} + j\sin{\theta}\,$ we can then differentiate z with respect to θ.

$\begin{align} \frac{dz}{d\theta} & = -\sin{\theta} + j\cos{\theta} \\ & = j^2\sin{\theta} + j\cos{\theta} \\ & = j(\cos{\theta} + j\sin{\theta}) \\ & = jz \end{align}$

The general solution to the differential equation $\frac{dz}{d\theta} = jz$ is $z = e^{j\theta+c}\,$.

This means that $\cos{\theta} + j\sin{\theta} = e^{j\theta+c}\,$

By putting θ as 0 we get:

$\begin{array}{rrcl} & \cos{0} + j\sin{0} & = & e^{0+c} \\ & 1 + 0j & = & e^{c} \\ \Rightarrow & c & = & 0 \end{array}$

So the general definition can be made:

$e^{j\theta} = \cos{\theta} + j\sin{\theta}\,$

For a complex number $z = x + yj\,$, calculating $e^z\,$ can be done:

$e^z = e^{x+yj} = e^xe^{yj} = e^x(\cos{y} + j\sin{y})\,$

Proof of de Moivre's theorem

We can now give an alternative proof of de Moivre's theorem for any rational value of n:

$\begin{align} (\cos{\theta} + j\sin{\theta})^n & = (e^{j\theta})^n \\ & = e^{jn\theta} \\ & = e^{j(n\theta)} \\ & = \cos{n\theta} + j\sin{n\theta} \\ \end{align}$

Complex roots

The roots of unity

The fundamental theorem of algebra states that a polynomial of degree n should have exactly n (complex) roots. This means that the simple equation zⁿ = 1 has n roots.

Let's take a look at z² = 1. This has two roots, 1 and -1. These can be plotted on an Argand diagram:

[Argand diagram]

Consider z³ = 1, from the above stated property, we know this equation has three roots. One of these is easily seen to be 1, for the others we rewrite the equation as z³ − 1 = 0 and use the factor theorem to obtain (z − 1)(z² + z + 1) = 0. From this, we can solve z² + z + 1 = 0 by completing the square on z so that we have $(z+\frac{1}{2})^2=-\frac{3}{4}$. Solving for z you obtain $z=-\frac{1}{2}\pm j\frac{\sqrt 3}{2}$. We have now found the three roots of unity of z³, they are z = 1, $z=-\frac{1}{2}+j\frac{\sqrt 3}{2}$ and $z=-\frac{1}{2}-j\frac{\sqrt 3}{2}$