A-level Mathematics/MEI/FP2/Complex Numbers

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 $$\theta$$ of its position vector in the Argand diagram.

[Argand diagram]

The distance r is the modulus of z, $$|z|\,$$. The angle $$\theta$$ is measured from the positive real axis and is taken anticlockwise. Adding any whole multiple of $$2\pi$$ 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:

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

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

Division
Dividing two complex number $$z_1\,$$ and $$z_2\,$$ in polar form:

$$z_1 = r_1(cos{\theta_1} + i\sin{\theta_1})\,$$

$$z_2 = r_2(cos{\theta_2} + i\sin{\theta_2})\,$$

⇒ $$ \frac{z_1}{z_2} = \frac{r_1(cos{\theta_1} + i\sin{\theta_1})}{r_2(cos{\theta_2} + i\sin{\theta_2})}$$

Multiply numerator and denominator by $$(cos{\theta_2} - i\sin{\theta_2})$$.

$$=\frac{r_1(cos{\theta_1} + i\sin{\theta_1})(cos{\theta_2} - i\sin{\theta_2})}{r_2(cos{\theta_2} + i\sin{\theta_2})(cos{\theta_2} - i\sin{\theta_2})}$$

Then, use distribution to simplify.

$$=\frac{r_1(cos{\theta_1}\cos{\theta_2} - i\cos{\theta_1}\sin{\theta_2} + i\sin{\theta_1}\cos{\theta_2} - i^2\sin{\theta_1}\sin{\theta_2})}{r_2(cos^2{\theta_2} - i\sin{\theta_2}\cos{\theta_2} + i\sin{\theta_2}\cos{\theta_2} - i^2\sin^2{\theta_2})}$$

Here, factorize by $$i$$ in the numerator and cancel out terms in the denominator. Note that $$i^2= -1$$.

$$\frac{r_1(cos{\theta_1}\cos{\theta_2} - i^2\sin{\theta_1}\sin{\theta_2} + i(sin{\theta_1}\cos{\theta_2} - cos{\theta_1}\sin{\theta_2})}{r_2(cos^2{\theta_2} + sin^2{\theta_2})}$$

$$=\frac{r_1(cos{\theta_1}\cos{\theta_2} + sin{\theta_1}\sin{\theta_2} + i(sin{\theta_1}\cos{\theta_2} - cos{\theta_1}\sin{\theta_2})}{r_2(1)}$$

Apply the formulas for the cosine of the difference of two angles and for the sine of the difference of two angles:

$$\frac{z_1}{z_2} = \frac{r_1}{r_2}(cos{(\theta_1 - \theta_2)} + i\sin{(\theta_1 - \theta_2)})$$.

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,

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

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

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

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 $$\theta$$ 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:

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

Summations
deMoivre's therom can come in handy for finding simple expressions for infinite series. This usually involves a series multiple angles, cosrθ, as in this next example:

Infinite series are defined by:

$$C=cos2\theta -\frac { 1 }{ 2 } cos5\theta+\frac{1}{4}cos8\theta -\frac{1}{8}cos11\theta +...$$

$$S=sin2\theta -\frac { 1 }{ 2 } sin5\theta+\frac{1}{4}sin8\theta -\frac{1}{8}sin11\theta +...$$

In order to find either the sum of C or the sum of S (or both!) you need to add C to jS: $$C+jS=cos2\theta +jsin2\theta -\frac { 1 }{ 2 } \left( cos5\theta +jsin5\theta \right) +\frac { 1 }{ 4 } \left( cos8\theta +jsin8\theta  \right) -\frac { 1 }{ 8 } \left( cos11\theta +jsin11\theta  \right) $$ Which using deMoivre's theorem can be written as:

$$C+jS=(cos\theta +jsin\theta )^{ 2 }-\frac { 1 }{ 2 } \left( cos\theta +jsin\theta \right)^5 +\frac { 1 }{ 4 } \left( cos\theta +jsin\theta  \right)^8 -\frac { 1 }{ 8 } \left( cos\theta +jsin\theta  \right)^{11}$$

It is easier to work with now using the form e^jθ:

$$C+jS=e^{ 2j\theta }-\frac { 1 }{ 2 } e^{ 5j\theta  }+\frac { 1 }{ 4 } e^{ 8j\theta  }+...$$

and you should be able to see the pattern, that the factor is negative 1/2 to the power of n-1 (the negative being alternately to even then odd powers is what makes it flip between + and -) and the power of e (the number in front of θ in our original equations) is equal to 3(n-1)jθ.

$$C+jS=-\left( \frac { -1 }{ 2 } \right) ^{ n-1 }e^{ (3n-1)j\theta  }=\left( \frac{-1}{2}e^{3j\theta}\right)^{n-1}$$

This is a geometric series, with a=

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 $$z^n = 1$$ has n roots.

Let's take a look at $$z^2 = 1$$. This has two roots, 1 and -1. These can be plotted on an Argand diagram:

[Argand diagram]

Consider $$z^3=1$$, 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 $$z^3-1=0$$ and use the factor theorem to obtain $$(z-1)(z^2+z+1)=0$$. From this, we can solve $$z^2+z+1=0$$ 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 $$z^3$$, they are $$z=1$$, $$z=-\frac{1}{2}+j\frac{\sqrt 3}{2}$$ and $$z=-\frac{1}{2}-j\frac{\sqrt 3}{2}$$

Solving an equation of the form $$ z^n = 1 $$
We know $$ z^6 = 1 $$ has six roots, one of which is 1.

We can rewrite this equation replacing the number 1 with $$ e^{2 \pi kj} $$ since 1 can be represented in polar form as having a modulus of 1 and an argument which is an integer multiple of $$ 2\pi $$.

$$ z^6 = e^{2 \pi kj} $$

Now by raising is side to the power of 1/6:

$$ z = e^{\frac{2 \pi kj}{6}} = e^{\frac{\pi kj}{3}} $$

To find all six roots we just change k, starting at 0 and going up to 5: