Distance Between Two Complex Numbers

Distance between two points and perpendicular bisectors are the two important terms in coordinate geometry as well as in complex numbers. It helps in finding the distance between two complex numbers. Measurement of distance is a very important aspect of our day-to-day life.This formula comes from the knowledge of basic distance formula and it's application..

JEE Main: Study Materials | High Scoring Topics | Preparation Guide

JEE Main: Syllabus | Sample Papers | Mock Tests | PYQs

This Story also Contains

1. Distance between two points
2. Perpendicular bisector
3. Equation of Circle
4. Equation of Circle in second form
5. Equation of Ellipse
6. Equation of Hyperbola
7. Section Formula
8. Centroid of Triangle

In this article, we will cover the concept of the distance between two points and perpendicular bisectors. This concept falls under the broader category of complex numbers and quadratic equations, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more.

Distance between two points

Distance between two points $\mathrm{A}\left({\mathrm{z}}_1\right)$ and $\mathrm{B}\left({\mathrm{z}}_2\right)$ is
$AB=|z_2-z_1|=\mid$ Affix of $B-$ Affix of $A\mid$

Let $z_1=x+iy$ and $z_2=x+iy$
Then, $|{\mathrm{z}}_1-{\mathrm{z}}_2|=|({\mathrm{x}}_1-{\mathrm{x}}_2)+\mathrm{i}({\mathrm{y}}_1-{\mathrm{y}}_2)|$ $=\sqrt{{({\mathrm{x}}_1-{\mathrm{x}}_2)}^2+{({\mathrm{y}}_1-{\mathrm{y}}_2)}^2}$
$=$ distance between points $({\mathrm{x}}_1,{\mathrm{y}}_1)$ and $({\mathrm{x}}_2,{\mathrm{y}}_2)=$ distance between between ${\mathrm{z}}_1$ and ${\mathrm{z}}_2$ where ${\mathrm{z}}_1={\mathrm{x}}_1+{\mathrm{iy}}_1$ and ${\mathrm{z}}_2={\mathrm{x}}_2+{\mathrm{iy}}_2$

• The distance of a point from the origin is $|z-0|=|z|$

• Three points $A\left(z_1\right),B\left(z_2\right)$ and $C\left(z_3\right)$ are collinear, then $AB+BC=AC$

i.e. $|z_2-z_1|+|z_3-z_2|=|z_3-z_1|$

Perpendicular bisector

We can use the distance formula to find the equation of perpendicular bisector
Let two fixed points $A\left(z_1\right)$ and $B\left(z_2\right)$ and a moving point $C(z)$ which lies on the perpendicular bisector of $AB$
As any point on the perpendicular bisector of $AB$ will be equidistant from $A$ and $B$, so

$\begin{array}{rl}& AC=BC\\ & |z-z_1|=|z-z_2|\end{array}$

This is the equation of perpendicular of the bisector of $AB$, where $A\left(z_1\right)$ and $B\left(z_2\right)$.

Equation of Circle

The equation of the circle whose center is at the point $z_0$ and has a radius $r$ is given by

$|z-z_0|=r$

If the center is the origin then, $z_0=0$, hence the equation reduces to $|z|=r$
Interior of the circle is represented by $|z-z_0|<r$
The exterior is represented by $|z-z_0|>r$
Here z can be represented as $\mathrm{x}+\mathrm{iy}$ and $z_0$ is represented by $x_0+i{y}_0$

Equation of Circle in second form

$\frac{|z-z_1|}{|z-z_2|}=k(k\ne 1,k>0)$

This equation also represents a circle. This can be verified by putting $z=x+iy,z_1=p+iq,z_2=a+ib$

Equation of Ellipse

$|z-z_1|+|z-z_2|=k,k>|z_1-z_2|$

|This represents an ellipse as the sum of distances of point $z$ from $z1$ and $z2$ is constant, which is the locus of an ellipse.

Equation of Hyperbola

$||\mathbf{z}-{\mathbf{z}}_1|-|\mathbf{z}-{\mathbf{z}}_2|\mid =\mathbf{k}(\mathrm{k}<|{\mathrm{z}}_1-{\mathrm{z}}_2|)$

This represents a hyperbola as the difference of distances of point $z$ from $z1$ and $z2$ is constant, which is the locus of a hyperbola.

Section Formula

The complex number $z$ dividing $z_1$ and $z_2$ internally in ratio $m:n$ is given by

$\mathrm{z}=\frac{m{z}_2+n{z}_1}{m+n}$

And
The complex number z dividing ${\mathrm{z}}_1$ and ${\mathrm{z}}_2$ externally in ratio $\mathrm{m}:\mathrm{n}$ is given by

$\mathrm{z}=\frac{m{z}_2-n{z}_1}{m-n}$

Centroid of Triangle

Centroid of the triangle with vertices $z_1,z_2$ and $z_3$ is given by $\frac{z_1+z_2+z_3}{3}$

Recommended Video Based on the Distance between two points and the perpendicular bisectors

Solved Examples Based On the Distance between two points and the Perpendicular bisectors

Example 1: Distance (in units) between $z_1=-3+2i$ and $z_2=-7-i$ equals
Solution:
As we learned in
Distance between Z 1 and Z 2 -
$|z_1-z_2|=|(x_1-x_2)+i(y_1-y_2)|=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}=$ Distance between points ${(x_1,y_1)}_{\mathrm{\&}}(x_2,y_2)=$ distance between $z_1$ and $z_2$
- wherein
$z_{1\mathrm{\&}}{z}_2$ are any two complex numbers, $z_1=x_1+i{y}_1,z_2=x_2+i{y}_2$
Distance between $Z_1$ and $Z_2=|Z_1-Z_2|$
Here, $Z_1-Z_2=4+3i$

$\therefore |Z_1-Z_2|=\sqrt{16+9}=5$

Hence, the answer is 5.

Example 2: If $z\ne 0$ and $\mathbf{z}$ moves such that $|z-1|=|z+1|$ then $|\arg (z)|\text{ equals }$
Solution:

As we learned in
Perpendicular bisector -
Locus of point equidistant from two given points.

$|z-z_1|=|z-z_2|$

z will lie on the perpendicular bisector of line joining $z_1$ and $z_2$.
- wherein
$z_1$ and $z_2$ are any two fixed points. $z$ is a moving point in the plain which is equidistant from $z_1$ and $z_2$. so $z$ will lie on the perpendicular bisector

$\because |Z-1|=|Z+1|\Rightarrow |Z-1|=|Z-(-1)|$

$\Rightarrow \mathrm{Z}$ lies on the perpendicular bisector of the line joining complex numbers
$Z_1=1$ and $Z_2=-1$, so the locus of $Z$ will be the imaginary axis.

$\Rightarrow \arg (Z)=\frac{\pi }{2},\frac{-\pi }{2}$

Hence, the answer is $\frac{\pi }{2}$.

Example 3: If $\left|\frac{1-iz}{z-i}\right|=1$ then the locus of $\mathbf{z}$ will be
1) Real axis
2) Imaginary axis
3) Argand plane
4) Circle

Solution: $◻$
As we learned in
Perpendicular bisector -
Locus of point equidistant from two given points.

$|z-z_1|=|z-z_2|$

z will lie on the perpendicular bisector of line joining $z_1$ and $z_2$.
- wherein

$z_1$ and $z_2$ are any two fixed points. z is a moving point in the plain which is equidistant from $z_1$ and $z_2.\mathrm{so}\mathrm{z}$ will lie on the perpendicular bisector

$\begin{array}{rl}& \left|\frac{1-iZ}{Z-i}\right|=1\Rightarrow |1-iZ|=|Z-1|\Rightarrow |i|\frac{1}{i}-Z|=|Z-i|\\ & \Rightarrow |-i-Z|=|Z-i|\Rightarrow |Z-(-i)|=|Z-i|\end{array}$

$\Rightarrow Z$ lies on the perpendicular bisector of the line joining i & -i
i.e real axis.

Hence, the answer is the option 1.

Example 4: If the set $\{\mathrm{Re}\left(\frac{z-\overline{z}+\overline{z}\overline{z}}{2-3z+5\overline{z}}\right):z\in C,\mathrm{Re}(z)=3\}$ is equal to the interval $(\alpha ,\beta ](\alpha ,\beta ]$, then $24(\beta -\alpha )$ is equal to
Solution:
Let ${}_1=\left(\frac{z-\overline{z}+z\overline{z}}{2-3z+5\overline{z}}\right)$
Letz $=3+iy$

$\overline{z}=3\text{-iy }$

$z_1=\frac{2\mathrm{iy}+(9+y^2)}{2-3(3+iy)+5(3-iy)}$
$=\frac{9+y^2+i(2y)}{8-8iy}$
$=\frac{(9+y^2)+i(2y)}{8(1-iy)}$
$\mathrm{Re}\left(z_1\right)=\frac{(9+y^2)-2y^2}{8(1+y^2)}$

$=\frac{9-y^2}{8(1+y^2)}$
$\begin{array}{rl}& =\frac{1}{8}\left[\frac{10-(1+y^2)}{(1+y^2)}\right]\\ & =\frac{1}{8}[\frac{10}{(1+y^2)}-1]\\ & 1+y^2\in [1,\mathrm{\infty }]\\ & \frac{1}{1+y^2}\in (0,1]\\ & \frac{10}{1+y^2}\in (0,10]\\ & \frac{10}{1+y^2}-1\in (-1,9]\\ & \mathrm{Re}\left(z_1\right)\in (\frac{-1}{8},\frac{9}{8}]\\ & \alpha =\frac{-1}{8},\beta =\frac{9}{8}\\ & 24(\beta -\alpha )=24(\frac{9}{8}+\frac{1}{8})=30\end{array}$

Hence, the answer is 30 .

Example 5: Let $z_1=2+3i$ and $z_2=3+4i$.The set $S=\{z\in C:{|z-z_1|}^2-{|z-z_2|}^2={|z_1-z_2|}^2\}\text{ represents a }$
1) hyperbola with the length of the transverse axis 7
2) hyperbola with eccentricity 2
3) a straight line with the sum of its intercepts on the coordinate axes equals -18
4) A straight line with the sum of its intercepts on the coordinate axes equals 14

Solution

$\begin{array}{rl}& \text{ Let }\mathrm{z}=\mathrm{x}+\mathrm{iy}\\ & \mathrm{z}-{\mathrm{z}}_1=(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}-3)\\ & {|\mathrm{z}-{\mathrm{z}}_1|}^2=(\mathrm{x}-2{)}^2+(\mathrm{y}-3{)}^2\\ & \mathrm{z}-{\mathrm{z}}_2=(\mathrm{x}-3)+\mathrm{i}(\mathrm{y}-4)\\ & {|\mathrm{z}-{\mathrm{z}}_2|}^2=(\mathrm{x}-3{)}^2+(\mathrm{y}-4{)}^2\\ & ((x-2{)}^2+(y-3{)}^2)-((x-3{)}^2+(y-4{)}^2)=2\\ & \Rightarrow 2\mathrm{x}+2\mathrm{y}=14\\ & =\mathrm{x}+\mathrm{y}=7\end{array}$

a straight line with the sum of intercept on $C.A=14$

Hence, the answer is the option 4.