# A-level Mathematics/MEI/FP2/Complex Numbers

< A-level Mathematics‎ | MEI‎ | FP2

## 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 ${\displaystyle \theta }$ of its position vector in the Argand diagram.

[Argand diagram]

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

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

[Argand diagram]

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

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

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

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

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


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

${\displaystyle \omega _{1}=r_{1}(\cos {\theta _{1}}+j\sin {\theta _{1}})\,}$

${\displaystyle \omega _{2}=r_{2}(\cos {\theta _{2}}+j\sin {\theta _{2}})\,}$

and then multiply them together,

${\displaystyle {\begin{array}{rl}\omega _{1}\omega _{2}&=r_{1}r_{2}(\cos {\theta _{1}}+j\sin {\theta _{1}})(\cos {\theta _{2}}+j\sin {\theta _{2}})\\&=r_{1}r_{2}((\cos {\theta _{1}}\cos {\theta _{2}}-\sin {\theta _{1}}\sin {\theta _{2}})+j(\sin {\theta _{1}}\cos {\theta _{2}}+\cos {\theta _{1}}\sin {\theta _{2}}))\\&=r_{1}r_{2}(\cos {(\theta _{1}+\theta _{2})}+j\sin {(\theta _{1}+\theta _{2})})\end{array}}\,}$

The result is a complex number with a modulus of ${\displaystyle r_{1}r_{2}\,}$ and an argument of ${\displaystyle \theta _{1}+\theta _{2}\,}$. This means that:

${\displaystyle |\omega _{1}\omega _{2}|=|\omega _{1}||\omega _{2}|\,}$

${\displaystyle \arg {(\omega _{1}\omega _{2})}=\arg {\omega _{1}}+\arg {\omega _{2}}\,}$

#### Division

Dividing two complex number ${\displaystyle z_{1}\,}$ and ${\displaystyle z_{2}\,}$ in polar form:

${\displaystyle z_{1}=r_{1}(cos{\theta _{1}}+i\sin {\theta _{1}})\,}$

${\displaystyle z_{2}=r_{2}(cos{\theta _{2}}+i\sin {\theta _{2}})\,}$

${\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}(cos{\theta _{1}}+i\sin {\theta _{1}})}{r_{2}(cos{\theta _{2}}+i\sin {\theta _{2}})}}}$

Multiply numerator and denominator by ${\displaystyle (cos{\theta _{2}}-i\sin {\theta _{2}})}$.

${\displaystyle ={\frac {r_{1}(cos{\theta _{1}}+i\sin {\theta _{1}})(cos{\theta _{2}}-i\sin {\theta _{2}})}{r_{2}(cos{\theta _{2}}+i\sin {\theta _{2}})(cos{\theta _{2}}-i\sin {\theta _{2}})}}}$

Then, use distribution to simplify.

${\displaystyle ={\frac {r_{1}(cos{\theta _{1}}\cos {\theta _{2}}-i\cos {\theta _{1}}\sin {\theta _{2}}+i\sin {\theta _{1}}\cos {\theta _{2}}-i^{2}\sin {\theta _{1}}\sin {\theta _{2}})}{r_{2}(cos^{2}{\theta _{2}}-i\sin {\theta _{2}}\cos {\theta _{2}}+i\sin {\theta _{2}}\cos {\theta _{2}}-i^{2}\sin ^{2}{\theta _{2}})}}}$

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

${\displaystyle {\frac {r_{1}(cos{\theta _{1}}\cos {\theta _{2}}-i^{2}\sin {\theta _{1}}\sin {\theta _{2}}+i(sin{\theta _{1}}\cos {\theta _{2}}-cos{\theta _{1}}\sin {\theta _{2}})}{r_{2}(cos^{2}{\theta _{2}}+sin^{2}{\theta _{2}})}}}$

${\displaystyle ={\frac {r_{1}(cos{\theta _{1}}\cos {\theta _{2}}+sin{\theta _{1}}\sin {\theta _{2}}+i(sin{\theta _{1}}\cos {\theta _{2}}-cos{\theta _{1}}\sin {\theta _{2}})}{r_{2}(1)}}}$

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

${\displaystyle {\frac {z_{1}}{z_{2}}}={\frac {r_{1}}{r_{2}}}(cos{(\theta _{1}-\theta _{2})}+i\sin {(\theta _{1}-\theta _{2})})}$.

## De Moivre's theorem

Using the multiplication rules we can see that if

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

then

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

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

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

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

## Complex exponents

### Definition

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

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

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

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

By putting ${\displaystyle \theta }$ as 0 we get:

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

So the general definition can be made:

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

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

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

### Proof of de Moivre's theorem

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

{\displaystyle {\begin{aligned}(\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{aligned}}}

### 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:

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

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

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

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

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

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

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.

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

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 ${\displaystyle z^{n}=1}$ has n roots.

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

[Argand diagram]

Consider ${\displaystyle z^{3}=1}$, 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 ${\displaystyle z^{3}-1=0}$ and use the factor theorem to obtain ${\displaystyle (z-1)(z^{2}+z+1)=0}$. From this, we can solve ${\displaystyle z^{2}+z+1=0}$ by completing the square on z so that we have ${\displaystyle (z+{\frac {1}{2}})^{2}=-{\frac {3}{4}}}$. Solving for z you obtain ${\displaystyle z=-{\frac {1}{2}}\pm j{\frac {\sqrt {3}}{2}}}$. We have now found the three roots of unity of ${\displaystyle z^{3}}$, they are ${\displaystyle z=1}$, ${\displaystyle z=-{\frac {1}{2}}+j{\frac {\sqrt {3}}{2}}}$ and ${\displaystyle z=-{\frac {1}{2}}-j{\frac {\sqrt {3}}{2}}}$

#### Solving an equation of the form ${\displaystyle z^{n}=1}$

We know ${\displaystyle z^{6}=1}$ has six roots, one of which is 1.

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

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

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

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

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