Title: Default for 10.2 The Argand Diagram and Polar form
Course: HELM Workbooks 2023-2024

The Argand Diagram

We know that a complex number is represented by \(z=x+i y\) in which \(x,y\) are real numbers and \(i^2={-1}\). We learned how to combine complex numbers together using the usual operations of addition, subtraction, multiplication and division. In this section we examine a useful geometrical description of complex numbers.

Since a complex number is specified by two real numbers \(x,y\) it is natural to represent a complex number by a vector in a plane. We take the usual \(Oxy\) plane in which the ‘horizontal’ axis is the \(x\)-axis and the ‘vertical’ axis is the \(y\)-axis.

[Image]

Thus the complex number \(z=2+3i\) would be represented by a line starting from the origin and ending at the point with coordinates \((2,3)\) and \(w=-1+i\) is represented by the line starting from the origin and ending at the point with coordinates \((-1,1)\). When the \(Oxy\) plane is used in this way it is called an Argand Diagram. With this interpretation the modulus of \(z\), that is \(|z|\) is the length of the line which represents \(z\).

Note: An alternative interpretation is to consider the complex number \(a+i b\) to be represented by the point \((a,b)\) rather than the line from \(0\) to \((a,b)\).

If we have two complex numbers \(z=a+i b\), \(w=c+i d\) then, as we already know \( z+w=(a+c)+i(b+d)\) that is, the real parts add together and the imaginary parts add together. But this is precisely what occurs with the addition of two vectors, which we will cover elsewhere but the process is quite natural.

If \(\underline{p}\) and \(\underline{q}\) are 2-dimensional vectors then: \[ \underline{p}=a \underline{i}+b \underline{j}\qquad \underline{q}=c \underline{i}+d \underline{j}\] where \(\underline{i}\) and \(\underline{j}\) are unit vectors in the \(x\)- and \(y\)-directions respectively. So, using vector addition: \[\underline{p}+\underline{q}=(a+c) \underline{i}+(b+d) \underline{j}\]

[Image]

We conclude from this that addition (and hence subtraction) of complex numbers is essentially equivalent to addition (subtraction) of two-dimensional vectors.

Because of this, complex numbers (when represented on an Argand diagram) are slidable — as long as you keep their length and direction the same, you can position them anywhere on an Argand diagram. We see that the Cartesian form of a complex number: \(z=a+i b\) is a particularly suitable form for addition (or subtraction) of complex numbers. However, when we come to consider multiplication and division of complex numbers, the Cartesian description is not the most convenient form that is available to us. A much more convenient form is the polar form which we now introduce.

The polar form of a complex number

We have seen that the complex number \(z=a+i b\) can be represented by a line pointing out from the origin and ending at a point with Cartesian coordinates \((a,b)\)

[Image]

To locate the point P we introduce polar coordinates \((r,\theta)\) where \(r\) is the positive distance from \(0\) and \(\theta\) is the angle measured from the positive \(x\)-axis, as shown in the figure above. From the properties of the right-angled triangle there is an obvious relation between \((a,b)\) and \((r,\theta )\): \[ a=r\cos\theta\qquad b=r\sin\theta\] or equivalently, \[ r=\sqrt{a^2+b^2}\qquad \tan\theta = \frac{b}{a}.\] This leads to an alternative way of writing a complex number: \[\begin{aligned} z=a+i b&=r\cos\theta +i r\sin\theta\\ &=r(\cos\theta +i\sin\theta)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\end{aligned}\]

The angle \(\theta\) is called the argument of \(z\) and written, for short, arg\((z)\). The non-negative real number \(r\) is the modulus of \(z\). We normally consider \(\theta\) measured in radians to lie in the interval \(-\pi<\theta\le \pi\) although any value \(\theta +2k\pi\) for integer \(k\) will be equivalent to \(\theta\). The angle \(\theta\) may be expressed in radians or degrees, it is standard to use radians and we should be explicit that we use degrees is this is the case.

Key Point

If \(z=a+i b\), \(a\neq 0\) then \[z=r(\cos\theta+i\sin\theta)\] in which \[r=|z|=\sqrt{a^2+b^2}\quad \text{and}\quad \tan(\theta) =\dfrac{b}{a}.\] Note to find \(\theta=\text{arg}(z)\) we need to know which quadrant \(z\) lies by examining the signs of \(a\) and \(b\). The inverse tan function will give an angle in the first or fourth quadrant so we need to add or subtract \(\pi\) if the complex number lies in the second or third quadrant.

Example

Find the polar coordinate form of

1. \(z=3+4i\)
2. \(z=-3-i\)

Solution

1. Here \(r=|z|=\sqrt{3^2+4^2}=\sqrt{25}=5\qquad \theta=\text{arg}(z)=\tan^{-1}({4\over 3})=0.93 \mathrm{rad}\) so that\(z=5(\cos 0.93 \mathrm{rad}+i\sin 0.93 \mathrm{rad}).\)
2. Here \(r=|z|=\sqrt{(-3)^2+(-1)^2}=\sqrt{10}\approx 3.16\qquad \theta=\text{arg}(z)=\tan^{-1}{(-1)\over {(-3)}}\) It is natural to assume that \(\tan^{-1}\dfrac{(-1)}{(-3)}=\tan^{-1}\left(\dfrac{1}{3}\right)\).
Using this value on your calculator (unless it is very sophisticated) you will obtain a value of about \(18.43^{\circ}\), which is approximately \(0.321\ \mathrm{radians}\), for \({\tan^{-1}({1\over 3})}\). This is incorrect since if we use the Argand diagram to plot \(z=-3-i\) we get:

[Image]

The angle \(\theta\) is clearly \(-180^{\circ}+18.43^{\circ}=-161.57^{\circ}=-2.82 \mathrm{rad}\). This example warns us to take care when determining arg\((z)\) purely using algebra. You will always find it helpful to construct the Argand diagram to locate the particular quadrant into which your complex number is pointing. Your calculator cannot do this for you. Finally, in this example, \(z=3.16\ (\cos -2.82 \mathrm{rad}+i \sin -2.82 \mathrm{rad})\).

Multiplication and Division of Complex Numbers in Polar form

Now we finally get to the true power of the polar form of a complex number. There is necessary effort in converting from cartesian form to polar form but then work makes multiplication and division much more convenient. To see why, let us consider two complex numbers in polar form: \[z=r(\cos\theta +i\sin \theta)\qquad w=t(\cos\phi +i\sin\phi ).\] Then the product \(zw\) is calculated in the usual way \[\begin{aligned} zw&=[r\ (\cos\theta +i\sin \theta)][t\ (\cos\phi +i\sin\phi )]\\ &\equiv rt\ [\cos\theta\cos\phi -\sin\theta\sin\phi +i(\sin\theta\cos\phi +\cos\theta\sin\phi)]\qquad\qquad\qquad\qquad\qquad\quad\\ &\equiv rt\ [\cos (\theta +\phi)+i\sin (\theta + \phi )]\end{aligned}\] in which we have used the standard trigonometric identities \( \cos (\theta +\phi )\equiv \cos\theta\cos\phi -\sin\theta\sin\phi\qquad \sin (\theta + \phi )\equiv\sin\theta\cos\phi +\cos\theta\sin\phi.\) We see that in calculating the product that the moduli \(r\) and \(t\) multiply together whilst the arguments \(\text{arg}(z)=\theta\) and \(\text{arg}(w)=\phi\) add together.

We see that in calculating the quotient that the moduli \(r\) and \(t\) divide whilst the arguments \(\text{arg}(z)=\theta\) and \(\text{arg}(w)=\phi\) subtract.

Key Point

If \(z=r(\cos\theta +i\sin\theta )\) and \(w=t(\cos\phi +i\sin\phi )\) then \[zw=rt(\cos (\theta +\phi)+i\sin (\theta + \phi ))\qquad {z\over w}={r\over t}(\cos (\theta -\phi) +i\sin (\theta -\phi ))\]

We conclude that addition and subtraction are most easily carried out in Cartesian form whereas multiplication and division are most easily carried out in polar form.

Complex Numbers and Rotations

We have seen that, when multiplying one complex number by another, the moduli multiply together and the arguments add together. If, in particular, \(w\) is a complex number with a modulus \(t\) that is \( w=t(\cos\phi +i\sin\phi) (\text{i.e.}\ r=t)\) and if \(z\) is a complex number with modulus 1 that is \( z=(\cos\theta +i\sin\theta )\) (i.e. \(r=1\)) then multiplying \(w\) by \(z\) gives \(wz=t(\cos (\theta +\phi ) +i\sin (\theta +\phi ))\)(using the Key Point above). We see that the effect of multiplying \(w\) by \(z\) is to rotate the line representing the complex number \(w\) anti-clockwise through an angle \(\theta\) which is arg\((z)\), and preserving the length.

[Image]

This result would certainly be difficult to obtain had we continued to use the Cartesian form. Since, in terms of the polar form of a complex number \(-1=1(\cos 180^{\circ}+i\sin 180^{\circ})\) we see that multiplying a number by \(-1\) produces a rotation through \(180^{\circ}\). In particular multiplying a number by \(-1\) and then by \((-1)\) again (i.e. \((-1)(-1)\)) rotates the number through \(180^{\circ}\) twice, totalling \(360^\circ\), which is equivalent to leaving the number unchanged. Hence the introduction of complex numbers has ‘ explained’ the accepted (though not obvious) result \((-1)(-1)=+1\).

Exercise

Consider the complex number \[z=[value] \] Evaluate the modulus of \(z\) \((r = |z|)\) and the principal value of the argument of \(z\ (\theta = \mathrm{Arg}(z)\)). Enter your results in the correct order, i.e. \([r,\theta]\).

\([r,\theta] = \)
Feedback: \[r=\sqrt{x^2+y^2}\] \[r=\sqrt{\left([value]\right)^2+\left([value]\right)^2}=\sqrt{[value]+[value]}\] \[r=\sqrt[value]=[value]\] \[\tan(\mathrm{arg}(z))=\frac{y}{x}=\frac[value][value]=[value]\] \[\mathrm{arg}(z)=[value] + k\cdot\pi\] Considering the appropriate quadrant of the complex plane to fix \(k\), we find that \[\mathrm{arg}(z)=[value]\].

Exercise

Consider the complex number \(z\) with modulus \(|z| = [value]\) and argument \(\displaystyle{\arg(z) = [value]} \).

Convert this number into the form \(p + qi\) where \(p\) and \(q\) are real. Your answer must be precise.
Feedback: \[p=r \cos(\theta)=[value]\cdot \cos([value])=[value]\ ,\] and, \[q=r \sin(\theta)=[value]\cdot \sin([value])=[value].\] Hence the answer is \([value]\).

Q: Exercise

Assume \(k>0\) is a real number.  For each complex number specify the argument.
- \(k+ki\) | 45 degrees
- \(k-ki\) | -45 degrees
- \(-k+ki\) | 135 degrees
- \(-k-ki\) | -135 degrees

Q: Exercise

Which of the following are true?
- \( |z^2| = |z|^2 \)
- \( z z^* = |z|^2 \)
- \( z z^* = z^2 \)
- \( z z^* = |z| \)
- \( |z^2| = z^2 \)
A: \( |z^2| = |z|^2 \); \( z z^* = |z|^2 \)
Feedback: Let \(z = r(\cos \theta + i\sin \theta)\). Then \(|z| = r \implies |z|^2 = r^2\). Also we have \(z^2 = r^2 (\cos \theta + i\sin \theta)^2\) which implies that \(|z^2| = r^2\). Therefore \( |z^2| = |z|^2 \). Let \(z = a + ib\). Then \(zz^* = (a+ib)(a-ib) = a^2 + b^2\). Note that \(|z| = \sqrt{a^2 + b^2}\) which implies that \(|z|^2 = a^2+b^2\). Therefore \( z z^* = |z|^2 \).

Exercise

If \( \displaystyle{z_1 = [value]}\) and \( \displaystyle{z_2 = [value]}\) find:
1. \( |z_1 z_2| = \)
2. \( \text{Arg}(z_1 z_2) = \)
Give your answers to \(3\) d.p.
Feedback: \( z_1 = [value] \) so \( |z_1| = [value] \) and \( \text{arg}(z_1) = [value]\) \( z_2 = [value] \) so \( |z_2| = [value] \) and \( \text{arg}(z_2) = [value]\) 1. \( |z_1 z_2| = |z_1||z_2| = [value] \) 2. \( \text{arg}(z_1 z_2) = \text{arg}(z_1) + \text{arg}(z_2) = [value] \) Have in mind that \( -\pi < \text{arg}(z) \leq \pi \) so you may need to add or subtract a \( \pi\) from your \( \text{arg} \) results.

Exercise

If \( \displaystyle{z_1 = [value]} \) and \(\displaystyle{ z_2 = [value] }\), evaluate in polar form:
1. \(\displaystyle{z_1 z_2 = }\)
2. \( \dfrac{z_1}{z_2} = \)
3. \( \dfrac{z_2}{z_1} = \)
Feedback: Our complex numbers are \( z_1 = [value]\) and \( z_2 = [value]\). 1. \( z_1 z_2 = [value] \) 2. \( \dfrac{z_1}{z_2} = [value]\) 3. \( \dfrac{z_2}{z_1} = [value] \)

Exercise

Express \(\displaystyle{z_1= [value]+[value]}\) and \( \displaystyle{z_2 =  [value]}\) in exponential form.

\( z_1 = \)

\( z_2 = \)

Your angle \( \theta\) should be a positive multiple of \( \pi\), don't write it as a floating point number!
Feedback: We know that \( z= r(\cos{\theta}+i\sin{\theta}) = re^{i\theta} .\) For \( z_1= \displaystyle [value]+[value]\) we have: \[ r_1 = \left|[value]+[value]\right|= \sqrt{([value])^2+([value])^2} = [value] \; \text{and} \; \textrm{arg}(z_1) = [value].\] Hence \( z_1= [value].\) For \( z_2= \displaystyle [value]\) we have: \[ r_1 = \left|[value]\right|= \sqrt{([value])^2+([value])^2} = [value] \; \text{and} \; \textrm{arg}(z_2) = [value].\] Hence \( z_2= [value].\)

Exercise

Express \( \displaystyle{z= [value] }\) in Polar form.

\( \displaystyle{z =} \)
Feedback: We know that \( z=a+ib= r(\cos{\theta}+i\sin{\theta}) = re^{i\theta}\), where \(\theta=\tan^{-1}(\frac{b}{a})\)and \(r=\sqrt{a^2+b^2}.\) For \( z\) we have: \( z = [value] = [value]( \cos[value] +i \sin[value] ).\)

Exercise

Given \( \displaystyle{z_1= [value] }\) and \(\displaystyle{z_2= [value]}\) find:
1. \( \displaystyle z= z_1^2 z_2 = \)
Write your answer in Cartesian form.
2. the modulus of \( z = \)
3. the argument of \(z = \)
Feedback: We are given \( \displaystyle{z_1= [value]} \) and \(\displaystyle{z_2= [value]}\). First, convert these into the polar form: We then have \(\displaystyle{z_1=[value]\left(\cos\left([value]\right)+i\sin\left([value]\right)\right)}\) and \(\displaystyle{z_2=[value]\left(\cos\left([value]\right)+i\sin\left([value]\right)\right)}\). 1. \(\displaystyle{z = z_1^2 z_2 = [value]}\). 2. The modulus of \(z\) is [value]. 3. The argument of \(z\) is \(\displaystyle[value]\).

Exercise

The Argand diagram shows the complex number \( z = [value] \) along with \(-z\), \(z^*\) and \(-z^*\).

[Image: Argand diagram with points A, B, C, D]

Which point is \( z \)?

Which point is \( z^* \)?

Which point is \( -z^* \)?
Feedback: \( z = [value] \) is represented by the point \( ([value],[value]) \) which is point [value]. \( z^* = [value] \) is represented by the point \( ([value],[value]) \) which is point [value]. \( -z^* = [value] \) is represented by the point \( ([value],[value]) \) which is point [value].

Exercise

Convert the following numbers to polar form by replacing r and theta by appropriate precise values:
1. \( z_1 = [value] = \)
2. \( z_2 = [value] = \)
Feedback: 1. \( z_1 = [value] \) is a real number so \( r = |z| = [value] \) and \( \theta = [value] \). So, we can write \(z_1\) in polar form as: \( z_1 = [value]\). 2. \( z_2 = [value] \) is an imaginary number so \( r = |z| = [value] \) and \( \theta = [value] \). So, we can write \(z_2\) in polar form as: \( z_2 = [value]\).

Exercise

If \( \displaystyle{z_1} = [value]\) and \( \displaystyle{z_2} = [value]\) find:
1. \(\displaystyle{ \left|\dfrac{z_1}{z_2}\right| =} \)
2. \( \displaystyle{\text{arg}\left(\dfrac{z_1}{z_2}\right) =} \)
Give your answers to \(3\) d.p.
Feedback: \( z_1 = [value] \) so \( |z_1| = [value] \) and \( \text{arg}(z_1) = [value]\) \( z_2 = [value] \) so \( |z_2| = [value] \) and \( \text{arg}(z_2) = [value]\) 1. \( \left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|} = [value] \) 2. \( \text{arg}\left(\dfrac{z_1}{z_2}\right) = \text{arg}(z_1) - \text{arg}(z_2) = [value] \) Have in mind that \( -\pi < \text{arg}(z) \leq \pi \) so you may need to add or subtract a \( \pi\) from your \( \text{arg} \) results.

Exercise

If \(z\) is a complex number with \(|z| = [value] \) and \(\displaystyle{ \text{arg}(z) = [value]}\), find the modulus and argument of \( \displaystyle {w = [value]} \).
1. \( |w| = \)
2. \( \text{arg}(w) = \)
Give an exact answer.
Feedback: We are given \(|z| = [value]\) and \(\displaystyle{ \text{arg}(z) = [value]}\). So \(z = |z| \left( \cos(\text{arg}(z)) + i\sin(\text{arg}(z)) \right) = [value] \) \( w = \displaystyle [value] = \dfrac{[value]+[value]}{[value]-[value]} = \displaystyle [value] \). So, 1. \( |w| = [value] \) 2. \( \text{arg}(w) = [value] \)

Exercise

Find the modulus and argument of \( z =\displaystyle \dfrac{([value])^[value] \cdot ([value])^[value]}{([value])^[value]} \)
1. \( |z| = \) (Give an exact answer)
2. \( \text{arg}(z) = \)  (Give your answer to 3 d.p.)
Feedback: \( z_1 = [value] \) so \( |z_1| = [value] \) and \( \text{arg}(z_1) = [value]\) \( z_2 = [value] \) so \( |z_2| = [value] \) and \( \text{arg}(z_2) = [value]\) \( z_3 = [value] \) so \( |z_3| = [value] \) and \( \text{arg}(z_3) = [value]\) 1. \( |z| = \dfrac{|z_1|^[value] |z_2|^[value]}{|z_3|^[value]} = \dfrac{([value]^[value]) \cdot ([value]^[value])}{[value]^[value]} = \displaystyle [value] \) 2. \( \text{arg}(z) = [value]\text{arg}(z_1) + [value]\text{arg}(z_2) - [value]\text{arg}(z_3) = [value] \) Have in mind that \( -\pi < \text{arg}(z) \leq \pi \) so you may need to add or subtract a \( \pi\) from your \( \text{arg} \) results.

Exercise

Express \( \displaystyle{z_1= [value] }\) and \( \displaystyle{z_2 = [value]}\) in Cartesian form.

\( z_1 = \)

\( z_2 = \)
Feedback: We know that \( z= r(\cos{\theta}+j\sin{\theta}) = re^{i\theta} .\) For \( z_1\) we have: \( z_1 = [value] = [value]( \cos[value] +i \sin[value] ) = [value].\) For \( z_2\) we have: \( z_2 = [value] = [value]( \cos[value] +i \sin[value] ) = [value].\)

Q: Exercise

If \(z=r(\cos\theta +i\sin\theta )\) and \(w=t(\cos\phi +i\sin\phi )\) find the polar expression for \({\dfrac{z}{w}}.\)
- \(\displaystyle{\frac{z}{w}=\frac{r}{t}(\cos(\theta-\phi)+i\sin(\theta-\phi))}\) (Correct!)
- \(\displaystyle{\frac{z}{w}=\frac{r}{t}(\cos(\theta+\phi)+i\sin(\theta+\phi))}\) (Try again! Be careful when computing this!)
- \(\displaystyle{\frac{z}{w}=\frac{r}{t}(\cos(\theta+\phi)+i\sin(\theta-\phi))}\) (Try again! Be careful when computing this!)
- \(\displaystyle{\frac{z}{w}=\frac{r}{t}(\cos(\theta-\phi)+i\sin(\theta+\phi))}\) (Try again! Be careful when computing this!)
A: \(\displaystyle{\frac{z}{w}=\frac{r}{t}(\cos(\theta-\phi)+i\sin(\theta-\phi))}\)
Feedback: \(z=r(\cos\theta+i\sin\theta)\) and \(w=t(\cos\phi +i\sin\phi )\). So, \[\frac{z}{w}=\frac{r(\cos\theta+i\sin\theta)}{t(\cos\phi +i\sin\phi )}=\frac{r(\cos\theta+i\sin\theta)}{t(\cos\phi +i\sin\phi )}\frac{(\cos\phi -i\sin\phi )}{(\cos\phi -i\sin\phi )}\\=\frac{r((\cos\theta\cos\phi+\sin\theta\sin\phi)+i(\sin\theta\cos\phi-\cos\theta\sin\phi))}{t}\\=\frac{r((\cos\theta\cos(-\phi)-\sin\theta\sin(-\phi))+i(\sin\theta\cos(-\phi)+\cos\theta\sin(-\phi)))}{t}=\frac{r(\cos(\theta-\phi)+i(\sin(\theta-\phi))}{t}.\]

Q: Given that \(z=1+\i\), \(w=\i\) , represent the three complex numbers \(z,w\) and \(2z-3w-1\) on an Argand diagram.
- [Image]
- [Image]
A: [Image]

Exercise

Given that \(z=[value]\), \(w=[value]\) , which of the points below represent \([value]z+[value]w\) on an Argand diagram.

[Image: Argand diagram with points A, B, C, D]

\([value]z+[value]w=\)
Feedback: We are given that \(z=[value]\) and \(w=[value]\). Then, \([value]z+[value]w=[value]([value])+[value]([value]).\) This is the point \([value]\) on the Argand diagram.