Distance Between Two Complex Numbers

Distance between two points

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

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

i.e. $\left|z_2-z_1\right|+\left|z_3-z_2\right|=\left|z_3-z_1\right|$

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 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 $z_1=-3+2 i$ and $z_2=-7-i$ equals
Solution:
As we 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 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: $\square$
As we 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.