Distance Between Two Complex Numbers

Distance Between Two Complex Numbers

Edited By Komal Miglani | Updated on Jul 02, 2025 08:01 PM IST

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..

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
Distance Between Two Complex Numbers
Distance Between Two Complex Numbers

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
$A B=\left|z_2-z_1\right|=\mid$ 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 $A B+B C=A C$

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Three collinear complex numbers

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

Perpendicular Bisector


$
\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 $x_0+i y_0$


Circle with radius r

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.


Frequently Asked Questions (FAQs)

1. What are complex numbers?

Complex numbers are the numbers in which complex or imaginary parts exist. It is represented as a+ib.

2. Write distance formula.


Let $z_1=x+$ iy 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 $\mathrm{z}_1=\mathrm{x}_1+\mathrm{i} \mathrm{y}_1$ and $\mathrm{z}_2=\mathrm{x}_2+\mathrm{iy}_2$

3. Write the equation of the perpendicular bisector.

$\left|z-z_1\right|=\left|z-z_2\right|$
This is the equation of perpendicular to the bisector of $A B$, where $A\left(z_1\right)$ and $B\left(z_2\right)$.

4. Write the equation of the circle.

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

$
\left|z-z_0\right|=r
$

5. Write the formula of the centroid of a triangle.

Centroid of the triangle with vertices $z_1, z_2$, and $z_3$ is given by $\left(z_1+z_2+z_3\right) / 3$.

6. What does the distance between two complex numbers represent geometrically?
The distance between two complex numbers represents the length of the straight line segment connecting the two points on the complex plane that correspond to these numbers. It's the shortest path between these two points in the complex plane.
7. How is the distance between two complex numbers calculated?
The distance between two complex numbers z1 = a + bi and z2 = c + di is calculated using the formula: |z1 - z2| = √[(a - c)² + (b - d)²]. This is similar to the distance formula in the real plane but uses the real and imaginary parts of the complex numbers.
8. Why can't we simply subtract two complex numbers to find their distance?
Simply subtracting two complex numbers gives us another complex number, not a real distance. We need to find the magnitude (absolute value) of this difference to get the actual distance, which is always a non-negative real number.
9. How is the distance formula for complex numbers related to the Pythagorean theorem?
The distance formula for complex numbers is a direct application of the Pythagorean theorem in the complex plane. The real part difference forms one side of a right triangle, the imaginary part difference forms the other, and the distance is the length of the hypotenuse.
10. Can the distance between two complex numbers ever be negative?
No, the distance between two complex numbers is always non-negative. It represents a physical length in the complex plane, which can never be negative.
11. How does the distance between complex numbers relate to vector operations?
The distance between two complex numbers is equivalent to the magnitude (length) of the vector obtained by subtracting one complex number from the other. This is why we use the absolute value notation in the formula |z1 - z2|.
12. What's the difference between the distance and the difference of two complex numbers?
The difference of two complex numbers is itself a complex number, while the distance is a non-negative real number representing the length of the line segment between the two numbers in the complex plane.
13. How does conjugation affect the distance between complex numbers?
Conjugation doesn't affect the distance between complex numbers. The distance between z1 and z2 is the same as the distance between their conjugates, z1* and z2*.
14. Is the distance between z1 and z2 always equal to the distance between z2 and z1?
Yes, the distance is commutative. The distance from z1 to z2 is always equal to the distance from z2 to z1, just like in real number line or in physical space.
15. How does scaling affect the distance between complex numbers?
If you multiply both complex numbers by the same real scalar k, the distance between them is multiplied by |k|. For example, the distance between 2z1 and 2z2 is twice the distance between z1 and z2.
16. Can two different pairs of complex numbers have the same distance between them?
Yes, many different pairs of complex numbers can have the same distance between them. For example, the distance between 0 and 1+i is the same as the distance between 3+2i and 4+3i.
17. How is the distance formula for complex numbers different from the distance formula on a real number line?
The distance formula for complex numbers is two-dimensional, considering both real and imaginary parts, while the distance formula on a real number line is one-dimensional, considering only the difference between two real numbers.
18. What happens to the distance when you add the same complex number to both z1 and z2?
Adding the same complex number to both z1 and z2 doesn't change the distance between them. This operation is equivalent to translating both points in the complex plane by the same amount.
19. How does the triangle inequality theorem apply to distances between complex numbers?
The triangle inequality theorem states that the distance between two complex numbers is always less than or equal to the sum of their distances from a third complex number. Mathematically, |z1 - z3| ≤ |z1 - z2| + |z2 - z3|.
20. Can the distance between two complex numbers ever be imaginary?
No, the distance between two complex numbers is always a real, non-negative number. It represents a physical length in the complex plane, which can't be imaginary.
21. How does the distance formula change if we use polar form of complex numbers?
If z1 = r1(cos θ1 + i sin θ1) and z2 = r2(cos θ2 + i sin θ2), the distance is given by √(r1² + r2² - 2r1r2 cos(θ1 - θ2)). This is derived from the law of cosines.
22. What's the relationship between the distance of two complex numbers and their moduli?
The distance between two complex numbers is always less than or equal to the sum of their moduli and greater than or equal to the absolute difference of their moduli. That is, ||z1| - |z2|| ≤ |z1 - z2| ≤ |z1| + |z2|.
23. How does the concept of distance extend to higher dimensional complex spaces?
In higher dimensional complex spaces (Cn), the distance is calculated using a generalization of the Pythagorean theorem, summing the squares of the differences of corresponding components: |z1 - z2| = √(|z11 - z21|² + |z12 - z22|² + ... + |z1n - z2n|²).
24. Why do we square the differences in the distance formula?
We square the differences to ensure that all terms are positive before adding them. This is necessary because the differences can be positive or negative, but distance must always be non-negative.
25. How is the distance between complex numbers related to the concept of a metric space?
The distance between complex numbers satisfies all the properties of a metric: non-negativity, identity of indiscernibles, symmetry, and the triangle inequality. Thus, the complex plane with this distance function forms a metric space.
26. Can the distance between two complex numbers ever be zero if the numbers are different?
No, the distance between two complex numbers is zero if and only if the numbers are identical. If z1 ≠ z2, then |z1 - z2| > 0.
27. How does rotation affect the distance between complex numbers?
Rotating both complex numbers by the same angle around the origin doesn't change the distance between them. This is because rotation preserves distances in the complex plane.
28. What's the geometric interpretation of |z1 - z2| = |z1| + |z2|?
This equation holds when z1 and z2 lie on the same ray from the origin. Geometrically, it means that the three points (0, z1, and z2) are collinear, with one point between the other two.
29. How is the distance formula for complex numbers related to the concept of norm?
The distance between two complex numbers z1 and z2 is equivalent to the norm (or magnitude) of their difference: |z1 - z2| = ||z1 - z2||. This connects the geometric concept of distance to the algebraic concept of norm.
30. Can the distance between two complex numbers be irrational?
Yes, the distance between two complex numbers can be irrational. For example, the distance between 0 and 1+i is √2, which is irrational.
31. How does the concept of distance relate to the argument of complex numbers?
While the distance doesn't directly involve the arguments of complex numbers, the difference in arguments can affect the distance. Two numbers with the same modulus but different arguments will have a non-zero distance between them.
32. What's the relationship between the distance formula and the complex conjugate?
The distance formula can be expressed using complex conjugates: |z1 - z2| = √((z1 - z2)(z1 - z2)*), where * denotes the complex conjugate. This formulation is equivalent to the standard distance formula.
33. How does the concept of distance in the complex plane relate to the concept of absolute value on the real line?
The distance between two complex numbers generalizes the concept of absolute difference on the real line. When dealing with real numbers only, |z1 - z2| reduces to the familiar |x1 - x2|.
34. Can the distance between two complex numbers ever be greater than the sum of their absolute values?
No, the distance between two complex numbers is always less than or equal to the sum of their absolute values. This is a consequence of the triangle inequality: |z1 - z2| ≤ |z1| + |z2|.
35. How does the distance formula change if we use the j operator (used in engineering) instead of i?
The distance formula remains the same whether we use i or j to represent the imaginary unit. The choice between i and j is merely a notational convention and doesn't affect the underlying mathematics.
36. What's the relationship between the distance of two complex numbers and the area of the parallelogram they form with the origin?
If z1 and z2 are two complex numbers, the area of the parallelogram formed by 0, z1, z2, and z1+z2 is given by |z1 × z2| = |Im(z1*z2)|. This is not directly related to the distance |z1 - z2|, but both involve the magnitudes of complex numbers.
37. How does the concept of distance between complex numbers extend to quaternions?
For quaternions q1 = a1 + b1i + c1j + d1k and q2 = a2 + b2i + c2j + d2k, the distance is calculated similarly: |q1 - q2| = √((a1-a2)² + (b1-b2)² + (c1-c2)² + (d1-d2)²). This is a four-dimensional generalization of the complex distance formula.
38. Can the distance between two complex numbers be used to define a circle in the complex plane?
Yes, a circle in the complex plane can be defined as the set of all points z such that |z - c| = r, where c is the center of the circle and r is its radius. This uses the concept of distance from a fixed point.
39. How does the distance formula relate to the concept of continuity in complex analysis?
The distance formula is crucial in defining continuity for complex functions. A function f(z) is continuous at a point z0 if for any ε > 0, there exists a δ > 0 such that |f(z) - f(z0)| < ε whenever |z - z0| < δ.
40. What's the relationship between the distance of complex numbers and their representation on the Riemann sphere?
The distance between complex numbers on the Riemann sphere is not the same as in the complex plane. The stereographic projection used to map the complex plane onto the Riemann sphere doesn't preserve distances, but it does preserve angles.
41. How does the concept of distance between complex numbers relate to the idea of a limit in complex analysis?
The concept of a limit in complex analysis is defined using distances. We say lim(z→z0) f(z) = L if for every ε > 0, there exists a δ > 0 such that |f(z) - L| < ε whenever 0 < |z - z0| < δ.
42. Can the distance formula be used to define neighborhoods in the complex plane?
Yes, an ε-neighborhood of a point z0 in the complex plane is defined as the set of all points z such that |z - z0| < ε. This uses the distance formula to create a circular region around z0.
43. How does the concept of distance relate to the conformal mapping property of complex functions?
While conformal mappings preserve angles, they generally don't preserve distances. However, they do preserve the ratios of infinitesimal distances, which is why they're said to be "locally" distance-preserving.
44. What's the relationship between the distance formula and the concept of an open set in complex analysis?
An open set in the complex plane is a set that contains an open disk around each of its points. This definition relies on the distance formula: for each point z in an open set, there exists an ε > 0 such that all points w with |w - z| < ε are also in the set.
45. How does the distance between complex numbers relate to the concept of a metric space in topology?
The complex plane with the distance function d(z1, z2) = |z1 - z2| forms a metric space. This means that the distance function satisfies the four metric space axioms: non-negativity, identity of indiscernibles, symmetry, and the triangle inequality.

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