Geometry of Complex Numbers | Argand Plane, Polar Form | Solved Examples, FAQs, Important Questions Free pdf download

📢 Master Geometry of Complex Numbers – Class 11 Mathematics 📚

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Are you preparing for Class 11 Board Exams, JEE Mains & Advanced? The Geometry of Complex Numbers is a crucial topic that bridges algebra and geometry, enhancing problem-solving skills for advanced mathematics.

The geometrical representation of a complex number is one of the fundamental laws of algebra. A complex number z = α + iβ can be denoted as a point P(α, β) in a plane called Argand plane, where α is the real part, and β is an imaginary part \(\begin{array}{l}\text{The value of i} = \sqrt{-1}\end{array} \).

🔢 What’s Inside This Chapter?
✅ Complex Numbers in the Argand Plane – Graphical Representation
✅ Modulus & Argument of a Complex Number – Polar Form
✅ Locus of Complex Numbers – Geometric Interpretation
✅ Rotation & Reflection using Complex Numbers – Applications in Geometry
✅ Shortcut Tricks & JEE-Level Problems

Representation of Complex Number Z Modulus on Argand Plane

Argand plane consists of the real axis (x-axis) and an imaginary axis (y-axis).

\(\begin{array}{l}\left| Z \right|=\sqrt{{{\left( \alpha -0 \right)}^{2}}+{{\left( \beta -0 \right)}^{2}}}\end{array} \)

\(\begin{array}{l}=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\end{array} \)

\(\begin{array}{l}=\sqrt{Re(z)^2 + Img(z)^2}\end{array} \)

\(\begin{array}{l}\left| z \right|=\left| \alpha +i\beta \right|=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}\end{array} \)

Conjugate of Complex Numbers on Argand Plane

A conjugate of a complex number is a number with the same real part and opposite sign of the imaginary part but equal in magnitude.

Let Z = α + iβ be the complex number.

The mirror image of Z = α + iβ along the real axis will represent the conjugate of the given complex number.

\(\begin{array}{l}Z=\alpha +i\beta\end{array} \) (complex number)

\(\begin{array}{l}\overline{Z}=\alpha -i\beta\end{array} \) (conjugate of complex number)

Distance between Two Points in Complex Plane

If two complex numbers Z₁ and Z₂ are denoted as P and Q, respectively, in the Argand plane, then the distance between P and Q will be given as,

\(\begin{array}{l}PQ=\left| {{z}_{2}}-{{z}_{1}} \right|\end{array} \)

\(\begin{array}{l}=\left| \left( {{\alpha }_{2}}-{{\alpha }_{1}} \right)+i\left( {{\beta }_{2}}-{{\beta }_{1}} \right) \right|\end{array} \)

\(\begin{array}{l}=\sqrt{{{\left( {{\alpha }_{2}}-{{\alpha }_{1}} \right)}^{2}}+{{\left( {{\beta }_{2}}-{{\beta }_{1}} \right)}^{2}}}\end{array} \)

For example, find the distance of a point P, Z = (3 + 4i) from the origin.

Solution:

|Z| = |3 + 4i|

\(\begin{array}{l}=\sqrt{{{3}^{2}}+{{4}^{2}}}\end{array} \)

= 5 units

Polar Form of Complex Number

Complex numbers can be represented in both rectangular and polar coordinates. Here, we will study the polar form of any complex number.

\(\begin{array}{l}Z=\left( \alpha +i\beta \right)\end{array} \)

Let ‘θ’ be the angle between ‘OP’ and the real axis.

\(\begin{array}{l}Z=\alpha +i\beta ,\,\,\,\left| z \right|=r\end{array} \)

OM = r cos θ = α

ON = r sin θ = β

Hence,

\(\begin{array}{l}Z=\alpha +i\beta\end{array} \)

= (OM) + i (ON)

\(\begin{array}{l}=r\cos \theta +i\,\,r\sin \theta\end{array} \)

\(\begin{array}{l}=r\left( \cos \theta +i\,\,\sin \theta \right)\end{array} \)

\(\begin{array}{l}r=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}=\left| z \right|=\left| \alpha +i\beta \right|\end{array} \)

Where ‘θ’ is called the argument of a complex number.

\(\begin{array}{l}\theta =\arg \left( z \right)\end{array} \)

For conjugate of a complex number,

\(\begin{array}{l}\arg \left( \overline{z} \right)=-\theta\end{array} \)

Algebra of Complex Numbers

Let’s discuss the different algebras of complex numbers. We mainly deal with addition, subtraction, multiplication and division of complex numbers.

Product of two complex numbers

\(\begin{array}{l}{{Z}_{1}}=\left( {{\alpha }_{1}}+i{{\beta }_{1}} \right)\end{array} \)

and

\(\begin{array}{l}{{Z}_{2}}=\left( {{\alpha }_{2}}+i{{\beta }_{2}} \right)\end{array} \)

in argand plane

\(\begin{array}{l}{{\theta }_{1}}=\arg \left( {{z}_{1}} \right)\end{array} \) , \(\begin{array}{l}{{\theta }_{2}}=\arg \left( {{z}_{2}} \right)\end{array} \) (given)

Now, on multiplying Z₁ Z₂ = Z

\(\begin{array}{l}Z=\left( {{\alpha }_{1}}+i{{\beta }_{1}} \right).\left( {{\alpha }_{2}}+i{{\beta }_{2}} \right)\end{array} \)

\(\begin{array}{l}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right).\,{{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right)\end{array} \)

\(\begin{array}{l}={{r}_{1}}{{r}_{2}}\left[ \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right]\end{array} \)

Let say

\(\begin{array}{l}{{r}_{1}}.\,{{r}_{2}}=r\end{array} \).

\(\begin{array}{l}Z=r\left( \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right)\end{array} \)

Division of two complex number

Let

\(\begin{array}{l}{{Z}_{1}}={{\alpha }_{1}}+i{{\beta }_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\,\sin {{\theta }_{1}} \right)\end{array} \)

\(\begin{array}{l}{{Z}_{2}}={{\alpha }_{2}}+i{{\beta }_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right)\end{array} \)

Where,

\(\begin{array}{l}{{\theta }_{1}}=\arg \left( {{Z}_{1}} \right)\end{array} \)

\(\begin{array}{l}{{\theta }_{2}}=\arg \left( {{Z}_{2}} \right)\end{array} \)

Now, Z represents a complex number

\(\begin{array}{l}Z=\frac{{{Z}_{2}}}{{{Z}_{1}}}={{Z}_{2}}Z_{1}^{-1}\end{array} \)

\(\begin{array}{l}Z={{Z}_{2}}Z_{1}^{-1}=\frac{{{Z}_{2}}\overline{{{Z}_{1}}}}{{{\left| Z \right|}^{2}}}\end{array} \)

\(\begin{array}{l}=\frac{{{r}_{2}}}{{{r}_{1}}}\left( \cos \left( {{\theta }_{2}}-{{\theta }_{1}} \right)+i\,\sin \left( {{\theta }_{2}}-{{\theta }_{1}} \right) \right)\end{array} \)

Note:

1. On the multiplication of two complex numbers, their argument is added.

\(\begin{array}{l}\theta ={{\theta }_{1}}+{{\theta }_{2}}\end{array} \)

2. On the division of two complex numbers, their argument is subtracted.

\(\begin{array}{l}\theta ={{\theta }_{1}}-{{\theta }_{2}}\end{array} \)

Equation of Straight Line Passing through Two Complex Points

In Algebra, we have studied that the equation of the straight line passing through two points (x₁, y₁) and (x₂, y₂) is

\(\begin{array}{l}y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)\end{array} \)

In complex numbers

Equation of straight line passes through two points A(Z₁), and B(Z₂) can be represented as,

\(\begin{array}{l}Z-{{Z}_{1}}=\frac{{{Z}_{2}}-{{Z}_{1}}}{\overline{{{Z}_{2}}}-\overline{{{Z}_{1}}}}\left( \overline{Z}-\overline{{{Z}_{1}}} \right)\end{array} \)

\(\begin{array}{l}\Rightarrow \frac{Z-{{Z}_{1}}}{{{Z}_{2}}-{{Z}_{1}}}=\frac{\overline{Z}-\overline{{{Z}_{1}}}}{\overline{{{Z}_{2}}}-\overline{{{Z}_{1}}}}\end{array} \)

\(\begin{array}{l}\text{Where}\ \overline{Z}\ \text{represent conjugate of Z}.\end{array} \)

Or

\(\begin{array}{l}\left| \begin{matrix} Z & \overline{Z} & 1 \\ {{Z}_{1}} & \overline{{{Z}_{1}}} & 1 \\ {{Z}_{2}} & \overline{{{Z}_{2}}} & 1 \\ \end{matrix} \right|=0\end{array} \)

Which is the required equation of the straight line.

Section Formula in Complex Numbers

Let A(Z₁), B(Z₂) and C(Z) be the three points on a line.

If C divides line AB (internally) in the ratio of m : n, so that

\(\begin{array}{l}\frac{AC}{BC}=\frac{m}{n}\end{array} \)

Then,

\(\begin{array}{l}Z=\frac{m\,{{Z}_{2}}+n\,{{Z}_{1}}}{m+n}\end{array} \)

Co-linearity condition

Let Z₁, Z₂ and Z₃ be the three points A(Z₁), B(Z₂) and C(Z₃).

So, the condition for collinearity will be,

|AB| + |BC| = |AC|

Or,

\(\begin{array}{l}\left| \begin{matrix} {{Z}_{1}} & \overline{{{Z}_{1}}} & 1 \\ {{Z}_{2}} & \overline{{{Z}_{2}}} & 1 \\ {{Z}_{3}} & \overline{{{Z}_{3}}} & 1 \\ \end{matrix} \right|=0\end{array} \)

Equation of Circle in Complex Plane

Equation of circle, whose centre is the point representing the complex number Z₀ and radius is ‘r’.

Equation of circle will be

\(\begin{array}{l}\left| Z-{{Z}_{0}} \right|=r\end{array} \).

Equation of circle if Z1 and Z2 be the diametric end points of the circle,

\(\begin{array}{l}\left( Z-{{Z}_{1}} \right)\left( \overline{Z}-\overline{{{Z}_{2}}} \right)+\left( Z-{{Z}_{2}} \right)\left( \overline{Z}-\overline{{{Z}_{1}}} \right)=0\end{array} \)

Solved Examples

Example 1: Find the length of the line segment joining the points −1−i and 2+3i.

Solution:

Since the coordinates in the complex plane are (2, 3) and (−1,−1), the required distance is 5.

We know that the distance between z₁ and z₂ is |z₁−z₂|. Therefore, the required length is |2+3i+1+i|=5.

Example 2: Let the complex numbers z₁, z₂ and z₃ be the vertices of an equilateral triangle. Let z₀ be the circumcentre of the triangle, then z₁² + z₂² + z₃² is equal to

Solution:

Let r be the circumradius of the equilateral triangle and ω the cube root of unity.

Let ABC be the equilateral triangle with z₁, z₂ and z₃ as its vertices A, B and C, respectively, with circumcentre O'(z₀).

\(\begin{array}{l}\text{The vectors}\ {O}’A,{O}’B,{O}’C\ \text{are equal and parallel to}\ O{A}’,O{B}’,O{C}’,\ \text{respectively}.\end{array} \)

  Then the vectors,

\(\begin{array}{l}\overrightarrow{O{A}’}={{z}_{1}}-{{z}_{0}}=r{{e}^{i\theta }}\\ \overrightarrow{O{B}’}={{z}_{2}}-{{z}_{0}}=r{{e}^{\left(\theta +\frac{2\pi }{3} \right)}}=r\omega {{e}^{i\theta }} \\\overrightarrow{O{C}’}={{z}_{3}}-{{z}_{0}}=r{{e}^{i\,\left(\theta +\frac{4\pi }{3} \right)}}\\=r{{\omega }^{2}}{{e}^{i\theta }} \\\ {{z}_{1}}={{z}_{0}}+r{{e}^{i\theta }},{{z}_{2}}={{z}_{0}}+r\omega {{e}^{i\theta }},{{z}_{3}}={{z}_{0}}+r{{\omega }^{2}}{{e}^{i\theta }} \\z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=3z_{0}^{2}+2(1+\omega +{{\omega }^{2}}){{z}_{0}}r{{e}^{i\theta }}+ (1+{{\omega }^{2}}+{{\omega }^{4}}){{r}^{2}}{{e}^{i2\theta }}\\ =3z_{^{0}}^{2},\end{array} \)

Since

\(\begin{array}{l}1+\omega +{{\omega }^{2}}=0=1+{{\omega }^{2}}+{{\omega }^{4}}\end{array} \).

Example 3: If the centre of a regular hexagon is at the origin and one of the vertexes on the Argand diagram is 1 + 2i, then find its perimeter.

Solution:

Let the vertices be z₀, z₁,.., z₅ with respect to centre O at origin and |z₀|= √5.

\(\begin{array}{l}\Rightarrow {{A}_{0}}{{A}_{1}}= |{{z}_{1}}-{{z}_{0}}|\,=\,|{{z}_{0}}{{e}^{i\,\theta }}-{{z}_{o}}| \\= |{{z}_{0}}||\cos \theta +i\sin \theta -1| \\=\sqrt{5}\,\sqrt{{{(\cos \theta -1)}^{2}}+{{\sin }^{2}}\theta } \\=\sqrt{5}\,\sqrt{2\,(1-\cos \theta )}\\=\sqrt{5}\,\,2\sin (\theta /2) \\{{A}_{0}}{{A}_{1}}=\sqrt{5}\,.\,2\sin \,\left(\frac{\pi }{6} \right)=\sqrt{5}\left( \text because \,\,\theta =\frac{2\pi }{6}=\frac{\pi }{3} \right)\end{array} \)

Similarly, 

\(\begin{array}{l}{{A}_{1}}{{A}_{2}}={{A}_{2}}{{A}_{3}}={{A}_{3}}{{A}_{4}}={{A}_{4}}{{A}_{5}}={{A}_{5}}{{A}_{0}}=\sqrt{5}\end{array} \)

Hence the perimeter of, regular polygon is

\(\begin{array}{l}={{A}_{o}}{{A}_{1}}+{{A}_{1}}{{A}_{2}}+{{A}_{2}}{{A}_{3}}+{{A}_{3}}{{A}_{4}}+{{A}_{4}}{{A}_{5}}+{{A}_{5}}{{A}_{0}}\\=\,\,6\sqrt{5}\end{array} \)

Frequently Asked Questions

Q1

What do you mean by the conjugate of a complex number?

If z = x+iy, then the conjugate of z is denoted by z̄ = x-iy. Conjugate will have the same real part and imaginary part with opposite signs but equal in magnitude.

Q2

Give the formula to find the distance between two points in the complex plane.

If two complex numbers are denoted as P(α₁, β₁) and Q(α₂, β₂), respectively, in the Argand plane, then the distance between P and Q is given by PQ = √[(α₂-α₁)² + (β₂– β₁)²].

Q3

How do you represent a complex number in polar form?

Let z = x+iy be the complex number. Then we use the formula x = r sin θ, y = r cos θ. We find r by using r = √(x²+y²). θ = tan^-1(y/x).
In polar form z = r cos θ + i sin θ. Categories Math Class 11, Complex Numbers

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✔️ Step-by-Step Explanations for Conceptual Clarity
✔️ Solved Examples & Extensive Practice Questions
✔️ Special Focus on JEE-Level Problems & Shortcuts

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📕 Written by: Neeraj Anand
🏛 Published by: ANAND TECHNICAL PUBLISHERS
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