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 $\theta$ 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,

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

Complex exponents

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=

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 $z^n = 1$ 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:

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