Argand Plane and Polar Representation: We know that an ordered pair of real numbers can be represented in the Cartesian plane. A Cartesian plane is a graph with two axes, one for the \(x\)-axis and one for the \(y\)-axis. These two axes are perpendicular to one another. The origin, \(O\) is located at the precise middle of the graph, where the two axes intersect.
Most practical disciplines that deal with geometry, such as astronomy, physics, engineering, and others, require Cartesian coordinates. This system is also helpful since you can plot a series of points that will give you answers to your equations. But how can we represent the complex numbers that have both real and imaginary parts? A complex number is represented on the Argand plane. A point \((x, y)\) in the argand plane represents a complex number of the type \(z=x+i y\).

Let’s learn more about the argand plane, representation of a complex number in polar form, modulus and argument of complex numbers.

Representing a Complex Number

A two-dimensional plane called the argand plane is used to represent a complex number. A complex number \(z=x+i y\) is represented as a point on the argand plane, which is analogous to a coordinate plane with \(x\) and \(y\) axes. A point in an argand plane may be written in polar form \((r, \theta)\), where \(r\) is the modulus of the complex number and \(\theta\) is the angle formed by the line connecting the point \((x, y)\) and the origin with respect to the positive \(x\)-axis.

Geometrical Representation of a Complex Number

A complex number \(z=x+iy\) can be represented by a point \((x, y)\) on the plane which is known as the argand plane. To represent \(z=x+i y\) geometrically we take two perpendicular lines \(X^{\prime} O X\) and \(Y^{\prime} O Y\) i.e., the real axis, and the imaginary axis respectively. Now, plot a point whose \(x\) and \(y\) coordinates are respectively the real and imaginary parts of \(z\). This point \(P(x, y)\) represents the complex number \(z=x+i y\).

If a complex number is purely real, then its imaginary part is zero. Therefore, it will be represented by a point on the \(x\)-axis.
A purely imaginary complex number will be a point on the \(y\)-axis.
Thus, there exists a one-one correspondence between the points of the plane and the members (elements) of the set \(\mathbb{C}\) of all complex numbers. This means that for every complex number \(z=x+i y\) there exists a unique point \((x, y)\) on the plane and vice versa.

Conjugate of a Complex Number

The mirror image of a complex number \(z\) with regard to the horizontal axis (or \(x\)-axis) is its complex conjugate. \(\bar{z}\) denotes the complex conjugate of the complex number \(z\). If \(z=x+i y\), then \(\bar{z}=x- iy\) is the complex conjugate of \(z\). If the complex number \(z\) is in the first quadrant, its image about the horizontal axis, the complex conjugate \(\bar{z}\), is in the fourth quadrant, as shown in the graphic below.

Modulus of a Complex Number

The distance between the origin and the point on the argand plane that represents the complex number \(z\) is called the modulus of that number.

Observe that the length of the line segment \(O P\) is called the modulus of \(z=x+i y\) and is denoted by \(|z|\).

Here, \(\triangle O M P\) is a right-angled triangle,

\(\therefore O P^{2}=O M^{2}+M P^{2}\)

\(\Rightarrow O P^{2}=x^{2}+y^{2}\)

\(\Rightarrow O P=\sqrt{x^{2}+y^{2}}\)

Thus, \(|z|=\sqrt{x^{2}+y^{2}}\)

\(\therefore \,|z| = \sqrt {{{\{ \operatorname{Re} (z)\} }^2} + {{\{ \operatorname{Im} (z)\} }^2}} \)

Argument of a Complex Number

The angle made by the directed line segment \(O P\) with the \(x\)-axis, and measured in the anti-clockwise direction is called the argument or amplitude of \(z=x+i y\).

Here, the angle \(\theta\) which \(O P\) makes with positive direction of \(x\)-axis is the argument or amplitude of \(z\). This is denoted by \(\arg (z)\) or \(\operatorname{amp} (z)\).

From the above figure, we have

\(\tan \theta  = \frac{{PM}}{{OM}} = \frac{y}{x} = \frac{{\operatorname{Im} (z)}}{{\operatorname{Re} (z)}}\)

\( \Rightarrow \theta  = {\tan ^{ – 1}}\left( {\frac{{Im\left( z \right)}}{{{Re} \left( z \right)}}} \right)\)

The unique value of \(\theta\), such that \(-\pi<\theta \leq \pi\), is called the principal value of the amplitude or principal argument.

The argument of \(z\) depends upon the quadrant in which the point \(P\) lies as discussed below.

Argument of z in Four Quadrants

• When \(x>0\) and \(y>0\)

Since \(x\) and \(y\) both are positive, \(P(x, y)\) representing \(z=x+i y\) lies in the first quadrant. Let \(\theta\) be the argument of \(z\) and let \(\alpha\) be the acute angle satisfying \(\tan \alpha=\left|\frac{y}{x}\right|\). Here, \(\theta=\alpha\).

• When \(x<0\) and \(y>0\)

Here, the point \(P(x, y)\) lies in the second quadrant. Let \(\theta\) be the argument of \(z\) and let \(\alpha\) be the acute angle satisfying the condition \(\tan \alpha=\left|\frac{y}{x}\right|\). Here \(\theta=\pi-\alpha\) as shown in the figure.

• When \(x<0\) and \(y<0\)

Here, the point \(P(x, y)\) lies in the third quadrant. Let \(\theta\) be the argument of \(z\) and let \(\alpha\) be the acute angle satisfying the condition \(\tan \alpha=\left|\frac{y}{x}\right|\).Then,

\(\theta=-(\pi-\alpha)\)
\(\therefore \theta=\alpha-\pi\)

• When \(x>0\) and \(y<0\)

Here, the point \(P(x, y)\) lies in the fourth quadrant. If \(\theta\) is the argument of \(z\) and \(\alpha\) is the acute angle satisfying the condition \(\tan \alpha=\left|\frac{y}{x}\right|\), then,

\(\theta=-\alpha\)

Vectorial Representation of a Complex Number

A complex number \(z=x+iy\) can be represented by the position vector \(O P\) of point \(P(x, y)\) in a two-dimensional plane because a complex number depends on two things:

(i) its modulus

(ii) its argument

In the figure given below, the complex number \(z=x+i y\) is represented by the vector \(\overrightarrow{O P}\). Here, \(|z|\) is the length \(O P\) and \(\arg (z)\) is the angle which the directed line \(O P\) makes with the positive direction of \(x\)-axis.

Polar Representation of a Complex Number

Let \(z=x+i y\) be a complex number represented by a point \(P(x, y)\) in the Argand plane.

Then, by the geometrical representation of \(z=x+i y\), we get,

\(O P=|z|\)

\(\angle P O X=\theta=\arg (z)\)

In \(\triangle P O M\), we get,

\(\cos \theta=\frac{O M}{O P}=\frac{x}{|z|}\)

\(\Rightarrow x=|z| \cos \theta\)

\(\sin \theta=\frac{P M}{O P}=\frac{y}{|z|}\)

\(\Rightarrow y=|z| \sin \theta\)

Since, \(z=x+i y\)

\(\Rightarrow z=|z| \cos \theta+i|z| \sin \theta\)

\(\therefore z=|z|(\cos \theta+i \sin \theta)\)

\(\Rightarrow z=r(\cos \theta+i \sin \theta)\)

Where,

\(r=|z|\)

\(\theta=\arg (z)\)

This form of \(z\) is called a polar form of \(z\). If we use the general value of the argument \(\theta\), then the polar form of \(z\) is given by

\(z=r[\cos (2 n \pi+\theta)+i \sin (2 n \pi+\theta)]\)

Where,

\(r=|z|\)

\(\theta=\arg (z)\)

\(n\) is an integer

Polar Representation in Four Quadrants

Let \(|z|=r\) and \(\alpha\) be the acute angle given by \(\tan \alpha  = \left| {\frac{{\operatorname{Im} (z)}}{{\operatorname{Re} (z)}}} \right|\).

Let \(\theta\) be the argument of \(z=x+i y\).

Condition	Quadrant	\(\theta\)	Polar Form
\(x>0,y>0\)	First	\(\theta=\alpha\)	\(z=r(\cos \alpha+i \sin \alpha)\)
\(x<0, y>0\)	Second	\(\theta=\pi-\alpha\)	\(z=r[\cos (\pi-\alpha)+i \sin (\pi-\alpha)]= r (-\cos \alpha +i \sin\alpha)\)
\(x<0, y<0\)	Third	\(\theta=-(\pi-\alpha)\)	\(z=r[\cos (\alpha-\pi)+i \sin (\alpha-\pi)]\Rightarrow z=r[-\cos \alpha-i \sin \alpha]\)
\(x>0, y<0\)	Fourth	\(\theta=-\alpha\)	\(z=r[\cos (-\alpha)+i \sin (-\alpha)]) =r(\cos \alpha-i \sin \alpha)\)

Solved Examples on Argand Plane and Polar Representation

Q.1. Find the modulus and argument of the following complex number. \(-2+2 i \sqrt{3}\)
Ans: Given: \(z=-2+2 i \sqrt{3}\)
Then, \(|z|=\sqrt{(-2)^{2}+(2 \sqrt{3})^{2}}=4\)
Let \(\alpha\) be the acute angle given by \(\tan \alpha = \left|{\frac{{\operatorname{Im} (z)}}{{\operatorname{Re} (z)}}} \right|\).
Then,
\(\tan \alpha=\left|\frac{2 \sqrt{3}}{-2}\right|=\sqrt{3}\)
\(\Rightarrow \alpha=\frac{\pi}{3}\)
Here, \(\operatorname{Re} (z) < 0\) and \(\operatorname{Im} (z) > 0\).
So, the point representing \(z\) lies in the second quadrant.
\(\therefore \arg (z)=\pi-\alpha\)
\(=\pi-\frac{\pi}{3}\)
\(\Rightarrow \arg (z)=\frac{2 \pi}{3}\)

Q.2. Find the modulus and argument of the following complex number. \(\frac{1+i}{1-i}\)
Ans: Given: \(z=\frac{1+i}{1-i}\)
Then,
\(z=\frac{1+i}{1-i} \times \frac{1+i}{1+i}\)
\(=\frac{(1+i)^{2}}{1-i^{2}}\)
\(=\frac{1+2 i+i^{2}}{1-i^{2}}\)
\(=\frac{1+2 i-1}{1+1}\)
\(=i\)
\(\therefore z=0+i\)
\(\Rightarrow|z|=\sqrt{0^{2}+1^{2}}=1\)
To find \(\arg (z)\),
Let \(\alpha\) be the acute angle given by \(\tan \alpha = \left|{\frac{{\operatorname{Im} (z)}}{{\operatorname{Re} (z)}}} \right|\).
Then, \(\tan \alpha=\frac{1}{0}=\infty\)
\(\Rightarrow \alpha=\frac{\pi}{2}\)
We find that \(\operatorname{Re} (z) = 0\) and \(\operatorname{Im} (z) = 1 > 0\)
So, the point representing \(z\) lies on \(y\)-axis.
\(\therefore \arg (z)=\alpha=\frac{\pi}{2}\)

Q.3. Write the following complex numbers in the polar form. \(1-i\)
Ans: Given: \(z=1-i\)
Then,
\(|z| = \sqrt {{{(\operatorname{Re} (z))}^2} + {{(\operatorname{Im} (z))}^2}} \)
\(=\sqrt{1^{2}+(-1)^{2}}\)
\(\therefore|z|=\sqrt{2}\)
Now, let \(\alpha\) be the acute angle given by \(\tan \alpha = \left|{\frac{{\operatorname{Im} (z)}}{{\operatorname{Re} (z)}}} \right|\).
Then,
\(\tan \alpha=\left|\frac{-1}{1}\right|=1\)
\(\Rightarrow \alpha=\frac{\pi}{4}\)
We find that the point \((1,-1)\) representing \(z\) lies in the fourth quadrant.
Therefore, the argument of \(z\) is given by,
\(\theta=-\alpha=-\frac{\pi}{4}\)
Hence, the polar form of \(z=1-i\) is:
\(r(\cos \theta+i \sin \theta)=\sqrt{2}\left\{\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right\}\)
\(\therefore r(\cos \theta+i \sin \theta)=\sqrt{2}\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)\)

Q.4. Put the complex number \(\frac{1+7 i}{(2-i)^{2}}\) in the form \(r(\cos \theta+i \sin \theta)\), where is a positive real number, and \(-\pi<\theta \leq \pi\).
Ans: Given: \(z=\frac{1+7 i}{(2-i)^{2}}\)
Then,
\(z=\frac{1+7 i}{4-4 i+i^{2}}\)
\(=\frac{1+7 i}{3-4 i}\)
Simplifying,
\(=\frac{1+7 i}{3-4 i} \times \frac{3+4 i}{3+4 i}\)
\(=\frac{-25+25 i}{25}\)
\(\Rightarrow z=-1+i\)
\(r=|z|\)
\(\therefore r=\sqrt{(-1)^{2}+(1)^{2}}\)
\(r=\sqrt{2}\)
Let \(\alpha\) be the acute angle given by \(\tan \alpha = \left|{\frac{{\operatorname{Im} (z)}}{{\operatorname{Re} (z)}}} \right|\).
Then,
\(\tan \alpha=\left|-\frac{1}{1}\right|=1\)
\(\Rightarrow \alpha=\frac{\pi}{4} .\)
The point \((-1,1)\) representing \(z\) lies in the second quadrant.
\(\therefore \theta=\arg (z)=\pi-\alpha\)
\(=\pi-\frac{\pi}{4}\)
\(\theta=\frac{3 \pi}{4}\)
Hence, \(z\) in the polar form is given by
\(z=\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)\)

Q.5. If for complex numbers \(z_{1}\) and \(z_{2}, \arg \left(z_{1}\right)-\arg \left(z_{2}\right)=0\), then show that \(\left|z_{1}-z_{2}\right|=|| z_{1}|-| z_{2}||\).
Ans: Let \(|z|=r_{1}\) and \(\left|z_{2}\right|=r_{2}\)
Given: \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)=0\)
i.e., \(\arg \left(z_{1}\right)=\arg \left(z_{2}\right)=\theta\)(say)
\(\therefore z_{1}=r_{1}(\cos \theta+i \sin \theta)\) and \(z_{2}=r_{2}(\cos \theta+i \sin \theta)\)
\(\Rightarrow z_{1}-z_{2}=\left(r_{1}-r_{2}\right) \cos \theta+i\left(r_{1}-r_{2}\right) \sin \theta\)
\(\Rightarrow\left|z_{1}-z_{2}\right|^{2}=\left(r_{1}-r_{2}\right)^{2} \cos ^{2} \theta+\left(r_{1}-r_{2}\right)^{2} \sin ^{2} \theta\)
\(\Rightarrow\left|z_{1}-z_{2}\right|^{2}=\left(r_{1}-r_{2}\right)^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)\)
\(\Rightarrow\left|z_{1}-z_{2}\right|^{2}=\left(r_{1}-r_{2}\right)^{2}\)
\(\Rightarrow\left|z_{1}-z_{2}\right|=\left|r_{1}-r_{2}\right|\)
\(\Rightarrow\left|z_{1}-z_{2}\right|=|| z_{1}|-| z_{2}||\)
Hence, proved.

Summary

A complex number \(z=x+iy\) can be represented by a point \((x, y)\) on the plane which is known as the Argand plane. It can also be represented by the position vector \(O P\) of point \(P(x, y)\). The polar representation of the same number is \(z=r(\cos \theta+i \sin \theta)\), where \(r=|z|\), and \(\theta=\arg (z)\).
The modulus of a complex number is the distance between the origin and the point on the argand plane that represents the complex number \(z\), and the angle made by the directed line segment \(O P\) which is from the origin \(O\) to the point \(P(x, y)\) that represents the complex number \(z=x+i y\) with the \(x\)-axis and measured in an anti-clockwise direction is called the argument or amplitude of \(z\).

Frequently Asked Questions (FAQs)

Q.1. What is meant by the Argand plane?
Ans: In a two-dimensional plane, the Argand plane is used to represent a complex number. A complex number \(z=x+i y\) is represented as a point on the argand plane analogous to a coordinate plane \((x, y)\). The real axis and imaginary axis of the argand plane are recognised as the \(x\)-axis and \(y\)-axis of the coordinate axis, respectively.

Q.2. How are complex numbers represented in the Argand plane?
Ans: A complex number \(z=a+i b\) can be represented by a point \((a, b)\) on the plane known as the Argand plane.
Step 1: Determine the complex number’s real and imaginary parts.
Step 2: Move as much as possible along the real axis.
Step 3: As much as possible, go parallel to the imaginary axis.
Step 4: On the Gaussian plane, the point you arrive at is the needed complex number representation.

Q.3. How do you represent polar form? 
Ans: The polar form of a complex number is: 
\(z=r(\cos \theta+i \sin \theta)\)
Where,
\(r=|z|=\sqrt{a^{2}+b^{2}}\)
\(a=r \cos \theta\)
\(b=r \sin \theta\)
\(\theta\) is the argument of \(z\)

Q.4. What is the polar form of a vector?
Ans: \((r, \theta)\) represents the polar form of a complex number, where the absolute value or modulus of the complex number is the length \(r\) of the vector in Polar form, and the angle \(\theta\) with the positive \(x\)-axis is the direction angle or argument of \(x+y i\).

Q.5. What is the conjugate of \(5 i\)?
Ans: Let \(z=5 i=0+i 5\)
We know that the conjugate of \(z=x+i y\) is \(\bar{z}=x-i y\)
\(\therefore\) Conjugate of \(z=\bar{z}=0-i 5=-5 i\)

Learn About Argument of Complex Numbers

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