Geometry of Complex Numbers

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. The value of i = 1\sqrt{-1}.

In this article, students will learn representation of Z modulus on Argand plane, polar form, section formula and many more. Solved examples and clear diagrams will help students to have a well understanding about the topic.

Table of Contents:

Representation of Z modulus on Argand Plane

Argand plane consists of real axis (x – axis) and imaginary axis (y – axis).
Complex Numbers on Argand Plane

Z=(α0)2+(β0)2\left| Z \right|=\sqrt{{{\left( \alpha -0 \right)}^{2}}+{{\left( \beta -0 \right)}^{2}}} =α2+β2=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}} =Re(z)2+Img(z)2=\sqrt{Re(z)^2 + Img(z)^2} z=α+iβ=α2+β2\left| z \right|=\left| \alpha +i\beta \right|=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}

Conjugate of Complex Numbers on argand plane

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

Let Z = α + iβ be the complex number.

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

Conjugate of Complex Numbers on argand plane

Z=α+iβZ=\alpha +i\beta (complex number)

Z=αiβ\overline{Z}=\alpha -i\beta (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 argand plane, then distance between P and Q will be given as,

Distance between Two Points in Complex Plane

PQ=z2z1PQ=\left| {{z}_{2}}-{{z}_{1}} \right| =(α2α1)+i(β2β1)=\left| \left( {{\alpha }_{2}}-{{\alpha }_{1}} \right)+i\left( {{\beta }_{2}}-{{\beta }_{1}} \right) \right| =(α2α1)2+(β2β1)2=\sqrt{{{\left( {{\alpha }_{2}}-{{\alpha }_{1}} \right)}^{2}}+{{\left( {{\beta }_{2}}-{{\beta }_{1}} \right)}^{2}}}

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


Distance between Two Points in Complex Plane Example

|Z| = |3 + 4i|

=32+42=5=\sqrt{{{3}^{2}}+{{4}^{2}}}=5 units

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Polar form of complex number

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

Z=(α+iβ)Z=\left( \alpha +i\beta \right)

Polar form of complex number

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

Z=α+iβ,z=rZ=\alpha +i\beta ,\,\,\,\left| z \right|=r

OM = r cos θ = α

ON = r sin θ = β

Hence, Z=α+iβZ=\alpha +i\beta

= (OM) + i (ON)

=rcosθ+irsinθ=r\cos \theta +i\,\,r\sin \theta =r(cosθ+isinθ)=r\left( \cos \theta +i\,\,\sin \theta \right) r=α2+β2=z=α+iβr=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}=\left| z \right|=\left| \alpha +i\beta \right|

Where ‘θ’ is called argument of complex number.

θ=arg(z)\theta =\arg \left( z \right)

For conjugate of complex number

arg(z)=θ\arg \left( \overline{z} \right)=-\theta

Complex Number Polar Form

Algebra of Complex Numbers

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

Product of two complex number

Let P and Q represent complex number

Z1=(α1+iβ1){{Z}_{1}}=\left( {{\alpha }_{1}}+i{{\beta }_{1}} \right) and

Z2=(α2+iβ2){{Z}_{2}}=\left( {{\alpha }_{2}}+i{{\beta }_{2}} \right) in argand plane

θ1=arg(z1){{\theta }_{1}}=\arg \left( {{z}_{1}} \right) θ2=arg(z2){{\theta }_{2}}=\arg \left( {{z}_{2}} \right) (given)

Algebra of Complex Number

Now, on multiplying Z1 Z2 = Z

Z=(α1+iβ1).(α2+iβ2)Z=\left( {{\alpha }_{1}}+i{{\beta }_{1}} \right).\left( {{\alpha }_{2}}+i{{\beta }_{2}} \right) =r1(cosθ1+isinθ1).r2(cosθ2+isinθ2)={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right).\,{{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right) =r1r2[cos(θ1+θ2)+isin(θ1+θ2)]={{r}_{1}}{{r}_{2}}\left[ \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right]

Let say r1.r2=r{{r}_{1}}.\,{{r}_{2}}=r.

Z=r(cos(θ1+θ2)+isin(θ1+θ2))Z=r\left( \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right)

Division of two complex number

Let Z1=α1+iβ1=r1(cosθ1+isinθ1){{Z}_{1}}={{\alpha }_{1}}+i{{\beta }_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\,\sin {{\theta }_{1}} \right) Z2=α2+iβ2=r2(cosθ2+isinθ2){{Z}_{2}}={{\alpha }_{2}}+i{{\beta }_{2}}={{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right)

Where, θ1=arg(Z1){{\theta }_{1}}=\arg \left( {{Z}_{1}} \right) θ2=arg(Z2){{\theta }_{2}}=\arg \left( {{Z}_{2}} \right)

Now, Z represent a complex number

Z=Z2Z1=Z2Z11Z=\frac{{{Z}_{2}}}{{{Z}_{1}}}={{Z}_{2}}Z_{1}^{-1} Z=Z2Z11=Z2Z1Z2Z={{Z}_{2}}Z_{1}^{-1}=\frac{{{Z}_{2}}\overline{{{Z}_{1}}}}{{{\left| Z \right|}^{2}}} =r2r1(cos(θ2θ1)+isin(θ2θ1))=\frac{{{r}_{2}}}{{{r}_{1}}}\left( \cos \left( {{\theta }_{2}}-{{\theta }_{1}} \right)+i\,\sin \left( {{\theta }_{2}}-{{\theta }_{1}} \right) \right)

Complex Number Algebra


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

θ=θ1+θ2\theta ={{\theta }_{1}}+{{\theta }_{2}}

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

θ=θ1θ2\theta ={{\theta }_{1}}-{{\theta }_{2}}

Rotation in Complex Number

Here will understand the concept of rotation in complex numbers.

Let ZA, ZB and ZC be the vertices of a triangle ΔABC

Rotation in Complex Number

AB=Z2Z1=ZABAC=Z3Z1=ZAC}Z3Z2=ZBC\left. \begin{matrix} AB={{Z}_{2}}-{{Z}_{1}}={{Z}_{AB}} \\ AC={{Z}_{3}}-{{Z}_{1}}={{Z}_{AC}} \\ \end{matrix} \right\}{{Z}_{3}}-{{Z}_{2}}={{Z}_{BC}}

Let, ZAB=r1eiθ1{{Z}_{AB}}={{r}_{1}}{{e}^{i{{\theta }_{1}}}} ZAC=r2eiθ2{{Z}_{AC}}={{r}_{2}}{{e}^{i{{\theta }_{2}}}} ZBC=r3eiθ3{{Z}_{BC}}={{r}_{3}}{{e}^{i{{\theta }_{3}}}}

So, that

ZABZAC=r1eiθ1r2eiθ2=r1r2ei(θ1θ2)\frac{{{Z}_{AB}}}{{{Z}_{AC}}}=\frac{{{r}_{1}}{{e}^{i{{\theta }_{1}}}}}{{{r}_{2}}{{e}^{i{{\theta }_{2}}}}}=\frac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i\left( {{\theta }_{1}}-{{\theta }_{2}} \right)}} =r1r2eiθ=\frac{{{r}_{1}}}{{{r}_{2}}}{{e}^{i\theta }} Z2Z1Z3Z1=Z2Z1Z3Z1eiθ=r1r2(cosθ+isinθ)\frac{{{Z}_{2}}-{{Z}_{1}}}{{{Z}_{3}}-{{Z}_{1}}}=\left| \frac{{{Z}_{2}}-{{Z}_{1}}}{{{Z}_{3}}-{{Z}_{1}}} \right|{{e}^{i\theta }}=\frac{{{r}_{1}}}{{{r}_{2}}}\left( \cos \theta +i\,\sin \theta \right)

Equation of Straight Line Passing through Two Complex Points

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

yy1=y2y1x2x1(xx1)y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)

In complex numbers

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

ZZ1=Z2Z1Z2Z1(ZZ1)Z-{{Z}_{1}}=\frac{{{Z}_{2}}-{{Z}_{1}}}{\overline{{{Z}_{2}}}-\overline{{{Z}_{1}}}}\left( \overline{Z}-\overline{{{Z}_{1}}} \right)b

ZZ1Z2Z1=ZZ1Z2Z1\Rightarrow \frac{Z-{{Z}_{1}}}{{{Z}_{2}}-{{Z}_{1}}}=\frac{\overline{Z}-\overline{{{Z}_{1}}}}{\overline{{{Z}_{2}}}-\overline{{{Z}_{1}}}}

Where Z\overline{Z} represent conjugate of Z


ZZ1Z1Z11Z2Z21=0\left| \begin{matrix} Z & \overline{Z} & 1 \\ {{Z}_{1}} & \overline{{{Z}_{1}}} & 1 \\ {{Z}_{2}} & \overline{{{Z}_{2}}} & 1 \\ \end{matrix} \right|=0

Which is the required equation of 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 ratio of m : n, so that


Equation of Straight Line Passing through Two Points

Then, Z=mZ2+nZ1m+nZ=\frac{m\,{{Z}_{2}}+n\,{{Z}_{1}}}{m+n}

Co-linearity condition

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

So, condition for collinearity will be

|AB| + |BC| = |AC|


Z1Z11Z2Z21Z3Z31=0\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

Equation of circle in Complex Plane

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

Equation of circle will be

Equation of circle in Complex Plane


ZZ0=r\left| Z-{{Z}_{0}} \right|=r.

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

(ZZ1)(ZZ2)+(ZZ2)(ZZ1)=0\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

Solved Examples

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


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

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

Example 2: Let the complex numbers z1,z2{{z}_{1}},{{z}_{2}} and z3{{z}_{3}} be the vertices of an equilateral triangle. Let z0{{z}_{0}} be the circumcentre of the triangle, then z12+z22+z32=z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=


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{{z}_{1}},{{z}_{2}}  and z3{{z}_{3}} as its vertices A,B and C respectively with circumcentre O(z0){O}'({{z}_{0}}). The vectors OA,OB,OC{O}’A,{O}’B,{O}’C are equal and parallel to OA,OB,OCO{A}’,O{B}’,O{C}' respectively. Then the vectors OA=z1z0=reiθOB=z2z0=re(θ+2π3)=rωeiθOC=z3z0=rei(θ+4π3)=rω2eiθ z1=z0+reiθ,z2=z0+rωeiθ,z3=z0+rω2eiθz12+z22+z32=3z02+2(1+ω+ω2)z0reiθ+(1+ω2+ω4)r2ei2θ=3z02,\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},

since 1+ω+ω2=0=1+ω2+ω41+\omega +{{\omega }^{2}}=0=1+{{\omega }^{2}}+{{\omega }^{4}}.

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


Let the vertices be z0,z1,..,z5{{z}_{0}},{{z}_{1}},…..,{{z}_{5}} with respect to centre O at origin and z0=5|{{z}_{0}}|\,=\sqrt{5}.

A0A1=z1z0=z0eiθzo=z0cosθ+isinθ1=5(cosθ1)2+sin2θ=52(1cosθ)=52sin(θ/2)A0A1=5.2sin(π6)=5(becauseθ=2π6=π3)\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)

Similarly, A1A2=A2A3=A3A4=A4A5=A5A0=5{{A}_{1}}{{A}_{2}}={{A}_{2}}{{A}_{3}}={{A}_{3}}{{A}_{4}}={{A}_{4}}{{A}_{5}}={{A}_{5}}{{A}_{0}}=\sqrt{5}

Hence the perimeter of, regular polygon is =AoA1+A1A2+A2A3+A3A4+A4A5+A5A0=65={{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}