# 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 = $\sqrt{-1}$.

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.

### Representation of Z modulus on Argand Plane

Argand plane consists of real axis (x – axis) and imaginary axis (y – axis).

$\left| Z \right|=\sqrt{{{\left( \alpha -0 \right)}^{2}}+{{\left( \beta -0 \right)}^{2}}}$,

$=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}$,

$=\sqrt{Re(z)^2 + Img(z)^2}$,

$\left| z \right|=\left| \alpha +i\beta \right|=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}$

### 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.

$Z=\alpha +i\beta$ (complex number)

$\overline{Z}=\alpha -i\beta$ (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,

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

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

Solution:

|Z| = |3 + 4i|

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

## 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.

$Z=\left( \alpha +i\beta \right)$

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

$Z=\alpha +i\beta ,\,\,\,\left| z \right|=r$

OM = r cos θ = α

ON = r sin θ = β

Hence, $Z=\alpha +i\beta$

= (OM) + i (ON)

$=r\cos \theta +i\,\,r\sin \theta$ $=r\left( \cos \theta +i\,\,\sin \theta \right)$ $r=\sqrt{{{\alpha }^{2}}+{{\beta }^{2}}}=\left| z \right|=\left| \alpha +i\beta \right|$

Where ‘θ’ is called argument of complex number.

$\theta =\arg \left( z \right)$

For conjugate of complex number

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

### 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

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

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

${{\theta }_{1}}=\arg \left( {{z}_{1}} \right)$,

${{\theta }_{2}}=\arg \left( {{z}_{2}} \right)$ (given)

Now, on multiplying Z1 Z2 = Z

$Z=\left( {{\alpha }_{1}}+i{{\beta }_{1}} \right).\left( {{\alpha }_{2}}+i{{\beta }_{2}} \right)$ $={{r}_{1}}\left( \cos {{\theta }_{1}}+i\sin {{\theta }_{1}} \right).\,{{r}_{2}}\left( \cos {{\theta }_{2}}+i\,\sin {{\theta }_{2}} \right)$ $={{r}_{1}}{{r}_{2}}\left[ \cos \left( {{\theta }_{1}}+{{\theta }_{2}} \right)+i\,\sin \left( {{\theta }_{1}}+{{\theta }_{2}} \right) \right]$

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

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

Division of two complex number

Let ${{Z}_{1}}={{\alpha }_{1}}+i{{\beta }_{1}}={{r}_{1}}\left( \cos {{\theta }_{1}}+i\,\sin {{\theta }_{1}} \right)$,

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

Where, ${{\theta }_{1}}=\arg \left( {{Z}_{1}} \right)$,

${{\theta }_{2}}=\arg \left( {{Z}_{2}} \right)$

Now, Z represent a complex number

$Z=\frac{{{Z}_{2}}}{{{Z}_{1}}}={{Z}_{2}}Z_{1}^{-1}$,

$Z={{Z}_{2}}Z_{1}^{-1}=\frac{{{Z}_{2}}\overline{{{Z}_{1}}}}{{{\left| Z \right|}^{2}}}$,

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

Note:

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

$\theta ={{\theta }_{1}}+{{\theta }_{2}}$

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

$\theta ={{\theta }_{1}}-{{\theta }_{2}}$

### 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

$y-{{y}_{1}}=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$

In complex numbers

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

$Z-{{Z}_{1}}=\frac{{{Z}_{2}}-{{Z}_{1}}}{\overline{{{Z}_{2}}}-\overline{{{Z}_{1}}}}\left( \overline{Z}-\overline{{{Z}_{1}}} \right)$,

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

Where $\overline{Z}$ represent conjugate of Z

Or

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

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

$\frac{AC}{BC}=\frac{m}{n}$

Then, $Z=\frac{m\,{{Z}_{2}}+n\,{{Z}_{1}}}{m+n}$

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|

Or,

$\left| \begin{matrix} {{Z}_{1}} & \overline{{{Z}_{1}}} & 1 \\ {{Z}_{2}} & \overline{{{Z}_{2}}} & 1 \\ {{Z}_{3}} & \overline{{{Z}_{3}}} & 1 \\ \end{matrix} \right|=0$

### 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

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

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

$\left( Z-{{Z}_{1}} \right)\left( \overline{Z}-\overline{{{Z}_{2}}} \right)+\left( Z-{{Z}_{2}} \right)\left( \overline{Z}-\overline{{{Z}_{1}}} \right)=0$

## Solved Examples

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

Solution:

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 is |2+3i+1+i|=5.

Example 2: Let the complex numbers ${{z}_{1}},{{z}_{2}}$ and ${{z}_{3}}$ be the vertices of an equilateral triangle. Let ${{z}_{0}}$ be the circumcentre of the triangle, then $z_{1}^{2}+z_{2}^{2}+z_{3}^{2}$ 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 ${{z}_{1}},{{z}_{2}}$  and ${{z}_{3}}$ as its vertices A,B and C respectively with circumcentre ${O}'({{z}_{0}})$. The vectors ${O}’A,{O}’B,{O}’C$ are equal and parallel to $O{A}’,O{B}’,O{C}'$ respectively. Then the vectors, $\overrightarrow{O{A}’}={{z}_{1}}-{{z}_{0}}=r{{e}^{i\theta }}\\ \overrightarrow{O{B}’}={{z}_{2}}-{{z}_{0}}=r{{e}^{\left(\theta +\frac{2\pi }{3} \right)}}=r\omega {{e}^{i\theta }} \\\overrightarrow{O{C}’}={{z}_{3}}-{{z}_{0}}=r{{e}^{i\,\left(\theta +\frac{4\pi }{3} \right)}}\\=r{{\omega }^{2}}{{e}^{i\theta }} \\\ {{z}_{1}}={{z}_{0}}+r{{e}^{i\theta }},{{z}_{2}}={{z}_{0}}+r\omega {{e}^{i\theta }},{{z}_{3}}={{z}_{0}}+r{{\omega }^{2}}{{e}^{i\theta }} \\z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=3z_{0}^{2}+2(1+\omega +{{\omega }^{2}}){{z}_{0}}r{{e}^{i\theta }}+ (1+{{\omega }^{2}}+{{\omega }^{4}}){{r}^{2}}{{e}^{i2\theta }}\\ =3z_{^{0}}^{2},$

since $1+\omega +{{\omega }^{2}}=0=1+{{\omega }^{2}}+{{\omega }^{4}}$.

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.

Solution:

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

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

Similarly, ${{A}_{1}}{{A}_{2}}={{A}_{2}}{{A}_{3}}={{A}_{3}}{{A}_{4}}={{A}_{4}}{{A}_{5}}={{A}_{5}}{{A}_{0}}=\sqrt{5}$

Hence the perimeter of, regular polygon is $={{A}_{o}}{{A}_{1}}+{{A}_{1}}{{A}_{2}}+{{A}_{2}}{{A}_{3}}+{{A}_{3}}{{A}_{4}}+{{A}_{4}}{{A}_{5}}+{{A}_{5}}{{A}_{0}}\\=\,\,6\sqrt{5}$

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