The Distance Between Two Complex Numbers
Let the complex number \( z_{1} = x_{1} \ \ + \ \ y_{1}i \) be represented by the point \( A(x_{1}, y_{1}) \) in the complex plane,
and let the complex number \( z_{2} = x_{2} \ \ + \ \ y_{2}i \) be represented by the point \( B(x_{2}, y_{2}) \).
The distance between these two complex numbers is given by:
\[
|AB| = |z_{1} – z_{2}| = \sqrt{ (x_{1} – x_{2})^{2} \ \ + \ \ (y_{1} – y_{2})^{2} }
\]
Example:
Find the distance between the complex numbers \( z_{1} = 2\sqrt{3} \ \ + \ \ 3i \) and \( z_{2} = \sqrt{3} \ \ + \ \ 2i \).
The complex number \( z_{1} \) is represented by \( A(2\sqrt{3}, 3) \), and \( z_{2} \) is represented by \( B(\sqrt{3}, 2) \).
Therefore, the distance is:
\[
|AB| = |z_{1} – z_{2}| = \sqrt{ (2\sqrt{3} – \sqrt{3})^{2} \ \ + \ \ (3 – 2)^{2} } = 2
\]
Example:
Sketch the graph of the set \( \{ z : |z \ – \ 5 \ – \ 6i| = 2, \ z \in \mathbb{C} \} \) in the complex plane.
Let \( z = x \ \ + \ \ yi \).
\[
|z – 5 – 6i| = 2
\Rightarrow |x + yi \ – \ 5 \ – \ 6i| = 2
\]
\[
\Rightarrow |(x \ – \ 5) + (y \ – \ 6)i| = 2
\]
\[
\Rightarrow \sqrt{ (x \ – \ 5)^{2} + (y \ – \ 6)^{2} } = 2
\]
\[
\Rightarrow (x \ – \ 5)^{2} + (y \ – \ 6)^{2} = 4
\]
Thus, the set of complex numbers \( z \) satisfying \( |z \ – \ 5 \ – \ 6i| = 2 \) is represented by a circle centered at \( M(5, 6) \) with a radius of \( R = 2 \) units.
Summary of Loci Results:
1. The locus of the set \( \{ z : |z \ – \ (a + bi)| = R, \ z \in \mathbb{C} \} \) in the complex plane is a circle centered at \( M(a, b) \) with radius \( R \).
2. The locus of the set \( \{ z : |z \ – \ (a + bi)| < R, \ z \in \mathbb{C} \} \) in the complex plane is the interior region of the circle centered at \( M(a, b) \) with radius \( R \).
3. The locus of the set \( \{ z : |z \ – \ (a + bi)| > R, \ z \in \mathbb{C} \} \) in the complex plane is the exterior region of the circle centered at \( M(a, b) \) with radius \( R \).
4. The locus of the set \( \{ z : R_{1} < |z \ – \ (a + bi)| < R_{2}, \ z \in \mathbb{C} \} \) in the complex plane is the annular region (annulus) bounded by concentric circles centered at \( M(a, b) \) with radii \( R_{1} \) and \( R_{2} \).
Example:
Find the locus of the set \( \{ z : |z + 1 \ – \ i| \leq 1, \ z \in \mathbb{C} \} \) in the complex plane.
\[
|z + 1 \ – \ i| \leq 1 \ \Rightarrow \ |z \ – \ (-1 + i)| \leq 1
\]
Since the complex constant \( z_{0} = -1 + i \) corresponds to the point \( M(-1, 1) \), the set of complex numbers \( z \) satisfying this inequality is represented by the circle centered at \( M(-1, 1) \) with a radius of \( R = 1 \) unit, including its interior.
Example:
Find the locus of the set \( \{ z : |z + 2i| > 1, \ z \in \mathbb{C} \} \) in the complex plane.
\[
|z + 2i| > 1 \ \Rightarrow \ |z – (0 – 2i)| > 1
\]
Since the complex constant \( z_{0} = 0 – 2i \) corresponds to the point \( M(0, -2) \), the solution set contains all points lying in the exterior region of the circle centered at \( M(0, -2) \) with radius \( R = 1 \) unit.
Example:
Find the locus of the set \( \{ z : 1 \leq |z – 3| < 2, \ z \in \mathbb{C} \} \) in the complex plane.
\[
1 \leq |z – 3| < 2
\Rightarrow
1 \leq |z – (3 + 0 \cdot i)| < 2
\]
Since the complex constant \( z_{0} = 3 + 0 \cdot i \) corresponds to the point \( M(3, 0) \), the graph of this system of inequalities is the region bounded between two concentric circles centered at \( M(3, 0) \) with radii \( R = 1 \) unit and \( R = 2 \) units, including the inner boundary but excluding the outer boundary.
Example:
Find the locus of the set \( \{ z : |2z| \geq |z – 3i|, \ z \in \mathbb{C} \} \) in the complex plane.
Let \( z = x + yi \).
\[
|2z| \geq |z – 3i|
\Rightarrow |2(x + yi)| \geq |x + yi – 3i|
\]
\[
\Rightarrow |2x + 2yi| \geq |x + (y – 3)i|
\]
\[
\Rightarrow \sqrt{ (2x)^{2} + (2y)^{2} } \geq \sqrt{ x^{2} + (y – 3)^{2} }
\]
\[
\Rightarrow 4x^{2} + 4y^{2} \geq x^{2} + y^{2} – 6y + 9
\]
\[
\Rightarrow 3x^{2} + 3y^{2} + 6y \geq 9
\]
\[
\Rightarrow x^{2} + y^{2} + 2y \geq 3
\]
\[
\Rightarrow x^{2} + (y + 1)^{2} \geq 2^{2}
\]
Thus, the locus satisfying this inequality consists of the circle centered at \( M(0, -1) \) with radius \( R = 2 \) units and its entire exterior region.
Example:
Find the locus of the set \( \{ z : \displaystyle\left| \frac{z \ – \ i }{z+1} \right| \leq 1, \ z \in \mathbb{C} \} \) in the complex plane.
*(Note: The typo in the original text fraction \( \frac{z-1}{z+1} \rightarrow \frac{z-i}{z+1} \) has been synchronized according to the math steps).*
Let \( z = x + yi \).
\[
\left| \frac{z – i}{z + 1} \right| \leq 1
\Rightarrow |z – i| \leq |z + 1|
\]
\[
\Rightarrow |x + yi – i| \leq |x + yi + 1|
\]
\[
\Rightarrow \sqrt{ x^{2} + (y – 1)^{2} } \leq \sqrt{ (x + 1)^{2} + y^{2} }
\]
\[
\Rightarrow x^{2} + y^{2} – 2y + 1 \leq x^{2} + 2x + 1 + y^{2}
\]
\[
\Rightarrow -2y \leq 2x
\]
\[
\Rightarrow y \geq -x
\]
The graph of the complex numbers \( z \) satisfying this inequality represents a closed half-plane bounded by the line \( y = -x \), as shown below:
Example:
Find the locus of the set \( \{ z : |z + i| \leq |z + 1| \ \text{ and } \ |z| \leq 2, \ z \in \mathbb{C} \} \) in the complex plane.
Let \( z = x + yi \).
\[
|z + i| \leq |z + 1| \
\]
\[
\Rightarrow |x + yi + i| \leq |x + yi + 1|
\]
\[
\Rightarrow \sqrt{ x^{2} + (y + 1)^{2} } \leq \sqrt{ (x + 1)^{2} + y^{2} }
\]
\[
\Rightarrow x^{2} + y^{2} + 2y + 1 \leq x^{2} + 2x + 1 + y^{2}
\]
\[
\Rightarrow y \leq x
\]
and
\[|z| \leq 2 \Rightarrow \sqrt{x^2 +y^2} \leq 2 \]
\[ x^2 + y^2 \leq 4 \]
The intersection of these regions corresponds to the section of a solid disk of radius 2 that lies below or on the line \( y = x \).
Example:
Find the locus of the set \( \{ z : |z + 4i| < |z| \ \text{ or } \ |z + i| \leq 1, \ z \in \mathbb{C} \} \) in the complex plane.
Let \( z = x + yi \).
From the first condition:
\[
|z + 4i| < |z|
\Rightarrow |x + yi + 4i| < |x + yi|
\]
\[
\Rightarrow \sqrt{ x^{2} + (y + 4)^{2} } < \sqrt{ x^{2} + y^{2} }
\]
\[
\Rightarrow x^{2} + y^{2} + 8y + 16 < x^{2} + y^{2}
\]
\[
\Rightarrow 8y < -16 \Rightarrow y < -2
\]
From the second condition:
\[
|z + i| \leq 1
\Rightarrow |z – (0 – i)| \leq 1
\]
This represents a disk centered at \( M(0, -1) \) with a radius of \( R = 1 \) unit, including its boundary. The combined union (“or”) forms the region shown below:
Example:
Find the locus of the set \( \{ z : |z + 1| \ – \ |z – 1| \geq 2, \ z \in \mathbb{C} \} \) in the complex plane.
Let \( z = x + yi \).
\[
|z + 1| \ – \ |z \ – \ 1| \geq 2
\Rightarrow |x + yi + 1| \ – \ |x + yi \ – \ 1| \geq 2
\]
\[
\Rightarrow \sqrt{ (x + 1)^{2} + y^{2} } \ – \ \sqrt{ (x \ – \ 1)^{2} + y^{2} } \geq 2
\]
Squaring both sides yields:
\[
x^{2} + 2x + 1 + y^{2}
\geq
4 + x^{2} \ – \ 2x + 1 + y^{2}
+ 4\sqrt{ (x \ – \ 1)^{2} + y^{2} }
\]
\[
\Rightarrow 4x – 4 \geq 4\sqrt{ (x – 1)^{2} + y^{2} }
\]
\[
\Rightarrow x \ – \ 1 \geq \sqrt{ x^{2} – 2x + 1 + y^{2} }
\]
Squaring both sides again under the constraint \( x \ – \ 1 \geq 0 \):
\[
x^{2} \ – \ 2x + 1 \geq x^{2} \ – \ 2x + 1 + y^{2}
\]
\[
\Rightarrow y^{2} \leq 0
\]
Since \( y \) must be a real number, \( y^{2} \leq 0 \) holds true only if \( y = 0 \). Combining this with the domain restriction \( x \geq 1 \), the locus represents a ray along the real axis where \( x \in [1, \infty) \).
Example:
Given \( z \in \mathbb{C} \) such that \( |z| \leq 3 \), find the maximum and minimum values of the expression \( |z \ – \ 8 \ – \ 6i| \).
The complex numbers satisfying \( |z| \leq 3 \) correspond to points on or inside a circle centered at the origin \( M(0, 0) \) with a radius of \( R = 3 \) units.
The geometric meaning of the expression \( |z \ – \ 8 \ – \ 6i| = |z – (8 + 6i)| \) is the distance between the variable point \( z \) and the fixed point \( z_{3} = 8 + 6i \).
The distance from the origin to \( z_{3} \) is \( \sqrt{8^2 + 6^2} = 10 \).
By choosing \( z = z_{1} \) along the line connecting the origin to \( z_3 \) on the opposite side of the circle, the distance reaches its maximum value:
\[ d_{\text{max}} = 10 + 3 = 13 \]
By choosing \( z = z_{2} \) on the boundary of the circle closest to \( z_3 \), the distance reaches its minimum value:
\[ d_{\text{min}} = 10 – 3 = 7 \]