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,

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

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.

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

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 $\theta$.

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

The general solution to the differential equation ${\frac {dz}{d\theta }}=jz$ 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:

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

Proof of de Moivre's theorem

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

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

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:

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

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

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: