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 $-\pi <\theta \le \pi$. 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,

$\begin{array}{cc}{\omega }_1{\omega }_2& =r_1{r}_2(\cos {\theta }_1+j\sin {\theta }_1)(\cos {\theta }_2+j\sin {\theta }_2)\\ & =r_1{r}_2((\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2)+j(\sin {\theta }_1\cos {\theta }_2+\cos {\theta }_1\sin {\theta }_2))\\ & =r_1{r}_2(\cos ({\theta }_1+{\theta }_2)+j\sin ({\theta }_1+{\theta }_2))\end{array}$

The result is a complex number with a modulus of $r_1{r}_2$ 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.

$=\frac{r_1(cos{\theta }_1\cos {\theta }_2-i\cos {\theta }_1\sin {\theta }_2+i\sin {\theta }_1\cos {\theta }_2-i^2\sin {\theta }_1\sin {\theta }_2)}{r_2(co{s}^2{\theta }_2-i\sin {\theta }_2\cos {\theta }_2+i\sin {\theta }_2\cos {\theta }_2-i^2{\sin }^2{\theta }_2)}$

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

$\frac{r_1(cos{\theta }_1\cos {\theta }_2-i^2\sin {\theta }_1\sin {\theta }_2+i(sin{\theta }_1\cos {\theta }_2-cos{\theta }_1\sin {\theta }_2)}{r_2(co{s}^2{\theta }_2+si{n}^2{\theta }_2)}$

$=\frac{r_1(cos{\theta }_1\cos {\theta }_2+sin{\theta }_1\sin {\theta }_2+i(sin{\theta }_1\cos {\theta }_2-cos{\theta }_1\sin {\theta }_2)}{r_2(1)}$

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 $\theta$.

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

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:

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

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:

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

Proof of de Moivre's theorem

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

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

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: $C+jS=cos2\theta +jsin2\theta -\frac{1}{2}(cos5\theta +jsin5\theta )+\frac{1}{4}(cos8\theta +jsin8\theta )-\frac{1}{8}(cos11\theta +jsin11\theta )$ Which using deMoivre's theorem can be written as:

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

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: