A-level Mathematics/MEI/FP2/Complex Numbers

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Modulus-argument form[edit]

Polar form of a complex number[edit]

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[edit]

The polar form of complex numbers can provide a geometrical interpretation of the multiplication and division of complex numbers.

Multiplication[edit]

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[edit]

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[edit]

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[edit]

Definition[edit]

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[edit]

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[edit]

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[edit]

The roots of unity[edit]

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 [edit]

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 6:

Applications of complex numbers in geometry[edit]