Distance Between Two Complex Numbers

Distance Between Two Complex Numbers

The distance between two complex numbers represents the length of the line segment joining their points on the Argand plane. It is calculated using the modulus of their difference and plays an important role in geometry, trigonometry, and complex number applications.

Distance Between Two Points in the Complex Plane

The distance between two points $A(z_1)$ and $B(z_2)$ is given by
$AB = |z_2 - z_1| = |\text{Affix of } B - \text{Affix of } A|$
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$. Then,
$|z_1 - z_2| = |(x_1 - x_2) + i(y_1 - y_2)| = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$
This is the same as the distance between points $(x_1, y_1)$ and $(x_2, y_2)$ on the Cartesian plane.

Distance from the Origin

The distance of a complex number $z$ from the origin is $|z - 0| = |z|$

Collinear Points in Complex Plane

Collinear points in the complex plane are points that lie on the same straight line when represented on the Argand diagram. The concept is useful in solving problems related to geometry, vectors, and proving properties of complex numbers.

Condition for Three Points to be Collinear

If three points $A(z_1)$, $B(z_2)$, and $C(z_3)$ are collinear, then
$|z_2 - z_1| + |z_3 - z_2| = |z_3 - z_1|$

This condition helps determine if three points lie on the same straight line in the complex plane.

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

Equations formed using Distance between two Complex Numbers

Equations formed using the distance between two complex numbers express geometric relationships on the complex plane. These equations use the distance formula |z_1-z_2| to represent curves, loci, or conditions involving complex points, helping solve problems in geometry and complex analysis effectively.

Equation of 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 $A B$
As any point on the perpendicular bisector of $A B$ will be equidistant from $A$ and $B$, so

$
\begin{aligned}
& A C=B C \\
& \left|z-z_1\right|=\left|z-z_2\right|
\end{aligned}
$

This is the equation of perpendicular of the bisector of $A B$, 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

$
\left|z-z_0\right|=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 $\left|z-z_0\right|<r$
The exterior is represented by $\left|z-z_0\right|>r$
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{\left|z-z_1\right|}{\left|z-z_2\right|}=k \quad(k \neq 1, k>0)
$

This equation also represents a circle. This can be verified by putting $z=x+i y, z_1=p+i q, z_2=a+i b$

Equation of Ellipse

$\left|z-z_1\right|+\left|z-z_2\right|=k, \quad k>\left|z_1-z_2\right|$

This represents an ellipse as the sum of distances of point $z$ from $z 1$ and $z 2$ is constant, which is the locus of an ellipse.

Equation of Hyperbola

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

This represents a hyperbola as the difference of distances of point $z$ from $z 1$ and $z 2$ is constant, which is the locus of a hyperbola.

Section Formula

The section formula helps find a complex number $z$ that divides the line segment joining two complex numbers $z_1$ and $z_2$ in a given ratio $m:n$.

• For internal division, the point zz lies between $z_1$ and $z_2$, and the formula is: $\mathrm{z}=\frac{m z_2+n z_1}{m+n}$

• For external division, the point zz lies outside the segment formed by $z_1$ and $z_2$, and the formula is: $\mathrm{z}=\frac{m z_2-n z_1}{m-n}$

These formulas are useful in geometry and complex number applications to locate points dividing segments proportionally, handling both cases where the point lies between or beyond the given points. The section formula extends naturally to problems involving division ratios, line segments, and coordinate calculations in the complex plane.

Centroid of Triangle

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

This is the point that acts like the average position of the triangle’s corners on the complex plane.

It’s where the three medians of the triangle meet, and each median is divided in the ratio 2:1 by the centroid. So, to find the centroid, just add the three vertex points as complex numbers and divide by 3—no need to separate real and imaginary parts explicitly.

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

Example 1: Distance (in units) between $z_1=-3+2 i$ and $z_2=-7-i$ equals

Solution:
As we have learned in,
Distance between $z_1$ and $z_2$,
$\left|z_1-z_2\right|=\left|\left(x_1-x_2\right)+i\left(y_1-y_2\right)\right|=\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}=$ Distance between points $\left(x_1, y_1\right)_{\&}\left(x_2, y_2\right)=$ distance between $z_1$ and $z_2$
- wherein
$z_{1 \&} z_2$ are any two complex numbers, $z_1=x_1+i y_1, z_2=x_2+i y_2$
Distance between $Z_1$ and $Z_2=\left|Z_1-Z_2\right|$
Here, $Z_1-Z_2=4+3 i$

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

Hence, the answer is 5.

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

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

$
\left|z-z_1\right|=\left|z-z_2\right|
$

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

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

$
\left|z-z_1\right|=\left|z-z_2\right|
$

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{aligned}
& \left.\left|\frac{1-i Z}{Z-i}\right|=1 \Rightarrow|1-i Z|=|Z-1| \Rightarrow|i| \frac{1}{i}-Z|=| Z-i \right\rvert\, \\
& \Rightarrow|-i-Z|=|Z-i| \Rightarrow|Z-(-i)|=|Z-i|
\end{aligned}
$

$\Rightarrow Z$ 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 $\left\{\operatorname{Re}\left(\frac{z-\bar{z}+\bar{z} \bar{z}}{2-3 z+5 \bar{z}}\right): z \in C, \operatorname{Re}(z)=3\right\}$ is equal to the interval $(\alpha, \beta](\alpha, \beta]$, then $24(\beta-\alpha)$ is equal to
Solution:
Let $_1=\left(\frac{z-\bar{z}+z \bar{z}}{2-3 z+5 \bar{z}}\right)$
Letz $=3+i y$

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

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

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

Hence, the answer is 30 .

Example 5: Let $z_1=2+3 i$ and $z_2=3+4 i$.The set $S=\left\{z \in C:\left|z-z_1\right|^2-\left|z-z_2\right|^2=\left|z_1-z_2\right|^2\right\} {\text { represents a }}$
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{aligned}
& \text { Let } \mathrm{z}=\mathrm{x}+\mathrm{iy} \\
& \mathrm{z}-\mathrm{z}_1=(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}-3) \\
& \left|\mathrm{z}-\mathrm{z}_1\right|^2=(\mathrm{x}-2)^2+(\mathrm{y}-3)^2 \\
& \mathrm{z}-\mathrm{z}_2=(\mathrm{x}-3)+\mathrm{i}(\mathrm{y}-4) \\
& \left|\mathrm{z}-\mathrm{z}_2\right|^2=(\mathrm{x}-3)^2+(\mathrm{y}-4)^2 \\
& \left((x-2)^2+(y-3)^2\right)-\left((x-3)^2+(y-4)^2\right)=2 \\
& \Rightarrow 2 \mathrm{x}+2 \mathrm{y}=14 \\
& =\mathrm{x}+\mathrm{y}=7
\end{aligned}
$

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

Hence, the answer is the option 4.

List of topics related to the Distance between two Complex Numbers

Below are topics related to the distance between two complex numbers that help build a stronger foundation in complex number geometry.

NCERT Resources

This section offers comprehensive NCERT solutions, exemplar problems, and revision notes for Class 11 Maths Chapter 5 on Complex Numbers and Quadratic Equations. These materials are designed to reinforce core concepts and support effective exam preparation.

NCERT Solutions for Class 11 Maths Chapter 5 -Complex Numbers and Quadratic Equations

NCERT Exemplar Class 11 Maths Solutions for Chapter 5 - Complex Numbers and Quadratic Equations

NCERT Class 11 Maths Notes for Chapter 5 - Complex Numbers and Quadratic Equations

Practice Questions based on the Distance between two Complex Numbers

Practice questions based on the distance between two complex numbers typically involve calculating the length of the line segment connecting points represented by complex numbers on the Argand plane. We have provided below the set of questions to practice Distance between two complex numbers:

Distance Formula And Equation Of Perpendicular Bisector - Practice Question MCQ

We have shared practice question sets for related topics below: