Interpretation of |z1-z2|

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 \({z_1} - {z_2}\):

Two fixed points in complex plane

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

Equal length vectors

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:

Distance between points

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:

Distance between variable and fixed points

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:

Variable and fixed points distance

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 \(\left( {2 - 3i} \right)\):

Variable and fixed points distance example 1

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:

Equi-distant 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:

Radii of concentric circles from origin

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More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus
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