Sometimes referred to as the Complex Plane, an “Argand Diagram” is where we can plot and illustrate complex numbers.

Because complex numbers of the form **a + b i**, can be represented as a vector.

## Argand Diagram Axis

On an Argand diagram, instead of an ** x** and

**axis, that we commonly see on a standard Cartesian Plane.**

*y*There is an

**Imaginary**axis, and a

**Real**axis.

__Example__Consider the complex number

**A**=

**2 + 3**.

*i*The REAL PART is 2, and the IMAGINARY PART is 3.

A can be drawn as a vector starting at the origin (0,0), and ending at the point (2,3).

Similarly, we can take the same approach with another complex number

**B**=

**-2**−

**2**.

*i*## Polar Form & Polar Coordinates,

Complex Number Modulus

A complex number can also be represented on the Argand diagram with what’s called Polar Coordinates.

The vector position would now be represented by

**(r,θ)**.

These values come from the POLAR FORM of a complex number.

Which is

**z**=

**r(cosθ**+

**, as opposed to**

*i*sinθ)**z**=

**a +**.

*i*bTo further understand this, consider a similar diagram to the one above.

Here

**r**is the modulus of

**z**, and is the magnitude/size of the vector. Generally denoted |

**z**|.

To establish r, we do the sum

**r**= \bf{\sqrt{a^2 \space + \space b^2}}.

θ is the angle the vector makes with the positive direction of the

**Real**axis. Usually in Radians, but can be in degrees also.

This is called the argument of

**z**, denoted Arg(

**z**).

From the triangle diagram above, using the Trig Ratios concerning θ.

cos

**θ**= \bf{\frac{a}{r}} sin

**θ**= \bf{\frac{b}{r}}

Multiplying both sides of both terms by

**r**gives.

**r**cos

**θ**=

**a**

**r**sin

**θ**=

**b**

So a complex number of the form

**z**=

**a +**, can become:

*i*b**z**=

**r**cos

**θ +**sin

*i*r**θ**=

**r**(cos

**θ**+

**sin**

*i***θ**).

This form can also be shortened to

**r c**.

*i*s θBut they represent the same thing.

For example,

**4(cos(**=

^{\bf{\frac{\pi}{2}}}) +*i*sin(^{\bf{\frac{\pi}{2}}}))**4 c**.

*i*s^{\bf{\frac{\pi}{2}}}## Polar Form Argument

As we’ve seen above, if we want to convert a complex number from the form

**z**=

**a +**into the form

*i*b**z**=

**r(cosθ +**.

*i*sinθ)Working out the modulus

**r**is straightforward.

[

**r**= \bf{\sqrt{a^2 \space + \space b^2}} ]

However, a little more care needs to be taken for the Argument θ.

If we refer to the Trigonometric Ratios for θ again. It’s the case that tan

**θ**= \bf{\frac{b}{a}}.

At first some people can sometimes assume that to obtain θ, we can always just do

tan

^{-1}(\bf{\frac{b}{a}}) =

**θ**.

But attention needs to be paid to which quadrant of the diagram the vector is actually in.

It’s the case that tan

^{-1}(\bf{\frac{b}{a}}) will produce a value between –

^{\bf{\frac{\pi}{2}}}and

^{\bf{\frac{\pi}{2}}}, or –90° and 90° in degrees.

Now this is fine for Quadrants 1 and 4.

But sometimes, a little extra step is needed to obtain an accurate Polar Form for quadrants 3 and 4. Depending on the values of a and b.

Let’s look at the variations of a complex number in the form

**z**=

**a +**.

*i*b### Case where a > 0:

**1)****1**+

*i*√3

**2)****1**−

*i*√3**1)**

**θ**= tan

^{-1}(\bf{\frac{\sqrt{3}}{1}}) = \bf{\frac{\pi}{3}}

**2)**

**θ**= tan

^{-1}(\bf{\frac{{\text{-}}\sqrt{3}}{1}}) = –\bf{\frac{\pi}{3}}

Case

**1)**is fine, that’s in the 1st quadrant.

Case

**2)**is also fine. The value of θ is in the 4th quadrant where the vector is.

### Case where a < 0:

**3)****-1**+

*i*√3

**4)****-1**−

*i*√3**3)**

**θ**= tan

^{-1}(\bf{\frac{\sqrt{3}}{{\text{-}}1}}) = –\bf{\frac{\pi}{3}}

**4)**

**θ**= tan

^{-1}(\bf{\frac{{\text{-}}\sqrt{3}}{{\text{-}}1}}) = \bf{\frac{\pi}{3}}

Now here, with

**a < 0**, we’re really to the left of the Imaginary axis, so a little extra step is required.

For 3), add to

*π*. =>

*π*+ -(\bf{\frac{\pi}{3}}) = \bf{\frac{2\pi}{3}}

For 4), add to –

*π*. => –

*π*+

^{\bf{\frac{\pi}{3}}}= –

^{\bf{\frac{2\pi}{3}}}

So putting vectors

**1)**,

**2)**,

**3)**and

**4)**into POLAR FORM.

The modulus

**r**is the same value each time, 2. As each vector is the same size/magnitude.

But the angle θ is different.

**1**+

**=>**

*i*√3**z**=

**2(cos(**+

^{\bf{\frac{\pi}{3}}})

*i*sin(^{\bf{\frac{\pi}{3}}}))**1**−

**=>**

*i*√3**z**=

**2(cos(-**+

^{\bf{\frac{\pi}{3}}})

*i*sin(-^{\bf{\frac{\pi}{3}}}))–

**1**+

**=>**

*i*√3**z**=

**2(cos(**

^{\bf{\frac{2\pi}{3}}}) +*i*sin(^{\bf{\frac{2\pi}{3}}}))–

**1**−

**=>**

*i*√3**z**=

**2(cos(-**+

^{\bf{\frac{2\pi}{3}}})

*i*sin(-^{\bf{\frac{2\pi}{3}}}))## Summary

A complex number can be in Cartesian Form, and Polar Form.

**CARTESIAN FORM**:

**z**=

**a + b**

*i***POLAR FORM**:

**z**=

**r(cosθ +**

*i*sinθ)When converting from Cartesian to Polar Form.

Make sure to pay attention to the a and b values.

**–**If

**a**

**> 0**,

**b**

**> 0**,

**θ**= tan

^{-1}(\bf{\frac{b}{a}})

**–**If

**a**

**> 0**,

**b**

**< 0**,

**θ**= tan

^{-1}(\bf{\frac{b}{a}})

**–**If

**a**

**< 0**,

**b**

**> 0**,

**θ**=

**+ tan**

*π*^{-1}(\bf{\frac{b}{a}})

**–**If

**a**

**< 0**,

**b**

**< 0**,

**θ**=

**–**+ tan

*π*^{-1}(\bf{\frac{b}{a}})

There is also the case where

**a**=

**0**,

**b**<

**0**or

**a**=

**0**,

**b**>

**0**.

When this happens, the vector on the Argand diagram will be either straight up, or straight down on the Imaginary axis.

So:

**–**If

**a**=

**0**,

**b**>

**0**,

**θ**=

^{\bf{\frac{\pi}{2}}}

**–**If

**a**=

**0**,

**b**<

**0**,

**θ**= –

^{\bf{\frac{\pi}{2}}}

**Return to TOP of page**