**COMPLEX NUMBERS**

**OVERVIEW**

**Below, we have provided our revision sheet for this topic, split into the following sections:**

**1) KEY POINTS**

**A summary of the most important aspects of the topic.**

**2) CAUSES OF CONFUSION**

**A list of mistakes that are commonly made by students.**

**3) EXAM TIPS**

**A collection of tips and tricks to give students that extra edge in their exam.**

**1) Key Points**

Any number in the form \(bi\), where \(i=\sqrt{-1}\) and \(b\in\mathbb{R}\), is know as an *imaginary number*.

Any number in the form \(a+bi\), where \(i=\sqrt{-1}\) and \(a,b\in\mathbb{R}\), is know as a *complex number*.

The complex number \(z=a+bi\) has a complex *conjugate* \(\bar{z}=a-bi\).

The *Conjugate Roots theorem* states that if \(z\) is a complex root of any polynomial equation of the form \(a_0x^n+a_1x^{n-1}+…+a_{n-1}x+a_n=0\), then \(\bar{z}\) is also a complex root of that equation.

*Cartesian* coordinates define a point on the \(xy\) plane by stating how far horizontally (\(x\) coordinate) and vertically (\(y\) coordinate) that point is from the origin.

*Polar* coordinates define a point on the \(xy\) plane by stating how far from the origin the point is (i.e. the *modulus *\(r\)) and how far rotated anticlockwise from the positive \(x\) axis the point is (the argument \(\theta\)).

For given Cartesian coordinates \((x,y)\), the corresponding modulus \(r\) is:

\begin{align}r=\sqrt{x^2+y^2}\end{align}

and the corresponding argument \(\theta\) is:

\begin{align}\theta&=A\mbox{ (First Quadrant)}\\\theta&=180^{\circ}-A\mbox{ (Second Quadrant)}\\\theta&=180^{\circ}+A\mbox{ (Third Quadrant)}\\\theta&=360^{\circ}-A\mbox{ (Fourth Quadrant)}\end{align}

where \(A=\tan^{-1}\left(\left|\dfrac{y}{x}\right|\right)\).

For given polar coordinates \((r,\theta)\), the corresponding Cartesian coordinates are:

\begin{align}x=r\cos \theta && y=r\sin\theta\end{align}

Any complex number \(a+bi\), \(a,b\in\mathbb{R}\), is represented on an *Argand Diagram* using Cartesian coordinates \((a,b)\).

If a complex number \(z_1=a+bi\) is added to another complex number \(z_2=c+di\) to obtain \(z_3=z_1+z_2\), \(z_3\) is equivalent to \(z_1\) being shifted \(c\) units horizontally and \(d\) units vertically on an Argand diagram.

If a complex number \(z_1=a+bi\) is multiplied by \(c\in\mathbb{R}\) to obtain \(z_2=cz_1\), the following is true:

- the
*modulus*of \(z_2\) is \(cr\), where \(r\) is the modulus of \(z_1\) - the
*argument*remains unchanged if \(c>0\) and increases by \(180^{\circ}\) is \(c<0\)

If a complex number \(z_1=a+bi\) is multiplied by \(i^n\), this causes \(z_1\) to rotate \(90^{\circ}\) *anticlockwise* \(n\) times on an Argand diagram.

If \(z_1\) is instead multiplied by \(-i^n\), this causes \(z_1\) to rotate \(90^{\circ}\) *clockwise* \(n\) times on an Argand diagram.

A complex number \(z\) can be written in *Cartesian form* \(z=a+bi\), where \(a\) and \(b\) describe how far horizontally and vertically \(z\) is from the origin on an Argand diagram.

A complex number \(z\) can also be written in *polar form* as \(z=r\cos\theta+r\sin\theta i\), where \(r\) and \(\theta\) are the modulus and argument of \(z\) on an Argand diagram.

If \(z=r\cos\theta+r\sin\theta i\), then:

\begin{align}\frac{1}{z}=\frac{1}{r}\cos(-\theta)+\frac{1}{r}\sin(-\theta)i\end{align}

where \(r\) and \(\theta\) are the modulus and argument of \(z\) respectively.

If \(z_1=r_1\cos\theta_1+r_1\sin\theta_1 i\) and \(z_2=r_2\cos\theta_2+r_2\sin\theta_2 i\), then:

\begin{align}z_1\times z_2=r_1r_2\cos(\theta_1+\theta_2)+r_1r_2\sin(\theta_1+\theta_2)i\end{align}

where \(r_1\) and \(\theta_1\) are the modulus and argument of \(z_1\) respectively and \(r_2\) and \(\theta_2\) are the modulus and argument of \(z_2\) respectively.

If \(z_1=r_1\cos\theta_1+r_1\sin\theta_1 i\) and \(z_2=r_2\cos\theta_2+r_2\sin\theta_2 i\), then:

\begin{align}\frac{z_1}{z_2}=\frac{r_1}{r_2}\cos(\theta_1-\theta_2)+\frac{r_1}{r_2}\sin(\theta_1-\theta_2)i\end{align}

where \(r_1\) and \(\theta_1\) are the modulus and argument of \(z_1\) respectively and \(r_2\) and \(\theta_2\) are the modulus and argument of \(z_2\) respectively.

*De Moivre’s theorem* states that if \(z=r\cos\theta+r\sin\theta i\), then:

\begin{align}z^n=r^n\cos(n\theta)+r^n\sin(n\theta)i\end{align}

where \(r\) and \(\theta\) are the modulus and argument of \(z\) respectively.

**2) ****Causes of Confusion**

A positive argument is created by rotating *anticlockwise* from the positive \(x\) axis.

A negative argument is instead created by rotating *clockwise* from the positive \(x\) axis.

Note that we use the positive \(x\) axis in *both* cases.

Often, we are able to find the argument quite easily as the angle within the triangle that we created happened to also be the argument itself.

However, this is not always the case!

To see this, consider the following four Cartesian points:

For each of these points, we can construct a right-angled triangle with one side on the \(x\) axis.

If we use Pythagoras’ theorem on each triangle, we find that the modulus, i.e. the length of each of the four blue lines, is \(\sqrt{3^2+4^2}=5\) in each case.

Instead, what makes each of these points have different polar coordinates is their argument.

To find each argument, we first find the angle \(A\) shown in each triangle below. This is what is known as the **reference angle**.

Recall that the argument is defined as the angle measured anticlockwise relative to the positive \(x\) axis. Therefore, in general, the reference angle \(A\) is not the argument \(\theta\)!

For all four triangles, however, the reference angle is found to be:

\begin{align}A&=\tan^{-1}\left(\frac{4}{3}\right)\\&\approx53.13^{\circ}\end{align}

How can we use these reference angles to find the corresponding argument?

Well, for the point \((3,4)\), i.e. the point in the **first quadrant**, the argument is simply the reference angle itself.

\begin{align}\theta&=A\\&=53.13^{\circ}\end{align}

as shown below.

For the point \((-3,4)\), i.e. the point in the **second quadrant**, the argument is instead given by:

\begin{align}\theta&=180^{\circ}-A\\&=180^{\circ}-53.13^{\circ}\\&=126.87^{\circ}\end{align}

as shown below.

For the point \((-3,-4)\), i.e. the point in the **third quadrant**, the argument is instead given by:

\begin{align}\theta&=180^{\circ}+A\\&=180^{\circ}+53.13^{\circ}\\&=233.13^{\circ}\end{align}

as shown below.

Finally, for the point \((3,-4)\), i.e. the point in the **fourth quadrant**, the argument is instead given by:

\begin{align}\theta&=360^{\circ}-A\\&=360^{\circ}-53.13^{\circ}\\&=306.87^{\circ}\end{align}

as shown below.

**3) ****Exam Tips**

Recall that purely real numbers are in fact types of complex numbers and can therefore be plotted on an Argand diagram (along the \(x\) axis).

Likewise, purely imaginary numbers can also be plotted on an Argand diagram (along the \(y\) axis).

If you blank in your exam on how to use De Moivre’s theorem to find *powers* of a complex number, you can also use the Binomial Theorem!

However, if you are instead asked to find the *roots* of a complex number, you *must* use De Moivre’s Theorem.