Interpretation of |z1-z2|

Let ${z_1}$ and ${z_2}$ represent two fixed points in the complex plane. There is a very useful way to interpret the expression $\left| {{z_1} - {z_2}} \right|$. Consider the following figure, which geometrically depicts the vector ${z_1} - {z_2}$:

However, observe that this vector is also equal to the vector drawn from the point ${z_2}$ to the point ${z_1}$:

Thus, $\left| {{z_1} - {z_2}} \right|$ represents the length of the vector drawn from ${z_2}$ to ${z_1}$. In other words, $\left| {{z_1} - {z_2}} \right|$ represents the distance between the points ${z_1}$ and ${z_2}$.

Let us take an example. Consider

$$\begin{align}&{z_1} = 1 + i\\&{z_2} =  - 3i\end{align}$$

The expression $\left| {{z_1} - {z_2}} \right|$, as we concluded, represents the distance between the points ${z_1}$ and ${z_2}$, which is $\sqrt {17}$, as is evident from the following figure:

We can verify this algebraically:

$$\begin{align}&{z_1} - {z_2} = \left( {1 + i} \right) - \left( { - 3i} \right) = 1 + 4i\\&\Rightarrow \,\,\,{z_1} - {z_2} = \sqrt {1 + 16}  = \sqrt {17} \end{align}$$

This interpretation of the expression $\left| {{z_1} - {z_2}} \right|$ as the distance between the points ${z_1}$ and ${z_2}$ is extremely useful and powerful. Let us see how.

Suppose that z is a variable point in the complex plane such that $\left| {z - i} \right| = 3$. What is the locus of z? In other words, what path does z trace out, while satisfying this constraint?

We can interpret $\left| {z - i} \right|$ as the distance between the variable point z and the fixed point i. The equation $\left| {z - i} \right| = 3$ says that the variable point z moves in such a way so that it is always at a constant distance of 3 units from the fixed point i. Thus, z traces out a circle in the plane, with center as the point i and radius 3 units:

Let’s take another example. Consider the equation

$$\left| {z - 1 + i} \right| = 2$$

We write this equation as

$$\left| {z - \left( {1 - i} \right)} \right| = 2$$

This says that the distance of z from the fixed point $\left( {1 - i} \right)$ is always 2 units. Thus, z traces out a circle in the plane, with center as the point $\left( {1 - i} \right)$ and radius equal to 2 units:

Example 1: z is a variable point in the plane such that

$$\left| {z - 2 + 3i} \right| = 11$$

Plot the locus of z.

Solution: We rewrite the given equation as

$$\left| {z - \left( {2 - 3i} \right)} \right| = 1$$

Thus, z traces out a circle of radius 1 unit, centered at the point $\left( {2 - 3i} \right)$:

Example 2: A variable point z always satisfies

$\left| {z - i} \right| = \left| {z + i} \right|$

As z moves, what path will it trace out in the plane?

Solution: First, we rewrite the given equation as

$$\left| {z - i} \right| = \left| {z - \left( { - i} \right)} \right|$$

This equation says that the distance of z from the point $i$ is equal to the distance of z from the point $\left( { - i} \right)$. Thus, z lies on the perpendicular bisector of these two points:

Clealy, z can lie anywhere on the real axis.

Example 3: Plot the region in which z can lie, if it satisfied $1 < \left| z \right| < 2$.

Solution: We can interpret $\left| z \right|$ or $\left| {z - 0} \right|$ as the distance between the point z and the origin. The given inequality says that the distance of the point z from the origin is greater than 1 but less than 2. Thus, z can lie anywhere in the following ring-shaped region: