Complex Numbers
14.0 Locus
14.0 Locus
1. Circle: The circle is represented by $$\left| {z - {z_0}} \right| = r$$ where ${z_0}$ is the centre of the circle and $r$ is the radius of the circle.
The equation of circle described on the line segment joining ${z_1}$ and ${z_1}$ as diameter is given by $${\text{Arg}}\left( {\frac{{z - {z_2}}}{{z - {z_1}}}} \right) = \pm \frac{\pi }{2}$$ or, $$(z - {z_1})(\overline z - \overline {{z_2}} ) + (z - {z_2})(\overline z - \overline {{z_1}} ) = 0$$
Note: The condition for the points ${z_1}$, ${z_2}$, ${z_3}$ and ${z_4}$ to be concyclic is $${\text{Arg}}\left( {\frac{{({z_2} - {z_3})({z_1} - {z_4})}}{{({z_1} - {z_3})({z_2} - {z_4})}}} \right) = \pm \pi ,0$$
2. Line segment: $${\text{Arg}}\left( {\frac{{z - {z_1}}}{{z - {z_2}}}} \right) = \theta $$
represents
- a line segment if $\theta = \pi $
- pair of ray if $\theta = 0$
- a part of circle if $0 < \theta < \pi $
3. Perpendicular bisector: The equation of the perpendicular bisector of the line segment joining ${z_1}$ and ${z_2}$ is $$z(\overline {{z_2}} - \overline {{z_1}} ) + \overline z ({z_2} - {z_1}) + {\left| {{z_1}} \right|^2} - {\left| {{z_2}} \right|^2} = 0$$
Proof: Consider a line segment joining $A({z_1})$ and $B({z_2})$ and a line $L$ is its perpendicular bisector. If $P(z)$ be any point on the $L$, we can say that $PA=PB$. Therefore, $$\begin{equation} \begin{aligned} \left| {z - {z_1}} \right| = \left| {z - {z_2}} \right| \\ \Rightarrow z(\overline {{z_2}} - \overline {{z_1}} ) + \bar z({z_2} - {z_1}) + {\left| {{z_1}} \right|^2} - {\left| {{z_2}} \right|^2} = 0 \\\end{aligned} \end{equation} $$
4. Ellipse: If $$\left| {z - {z_1}} \right| + \left| {z - {z_2}} \right| = 2a$$ where $2a > \left| {{z_1} - {z_2}} \right|$ then, point $z$ describes an ellipse whose foci are at ${z_1}$ and ${z_2}$ and $a \in {R^ + }$
5. Hyperbola: If $$\left| {z - {z_1}} \right| + \left| {z - {z_2}} \right| = 2a$$ where $2a < \left| {{z_1} - {z_2}} \right|$ then, point $z$ describes a hyperbola whose foci are at ${z_1}$ and ${z_2}$ and $a \in {R^ + }$
Question 13. Find the locus of $z$ for the following:
(i) $\left| {z - 3 + 2i} \right| - \left| {z + i} \right| = 0$
(ii) ${\text{Arg}}\left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{\pi }{4}$
(iii) $\left| {z - 1} \right| + \left| {z + i} \right| = 10$
(iv) $\left| {z - 3 + i} \right| - \left| {z + 2 - i} \right| = 1$
Solution: (i) We can write the form given in the question as $$\left| {z - 3 + 2i} \right| = \left| {z + i} \right|$$ which is same as $$\left| {z - {z_1}} \right| = \left| {z - {z_2}} \right|$$ From the above discussion on locus, we can say that locus of $z$ represents the perpendicular bisector of the line segment joining the points ${z_1} = 3 - 2i$ and ${z_2} = - i$.
(ii) Compare the given equation in the question with $${\text{Arg}}\left( {\frac{{z - {z_1}}}{{z - {z_2}}}} \right) = \theta $$ Since $$0 < \theta = \frac{\pi }{4} < \pi $$ Therefore, the locus of $z$ represents part of circle (major arc).
(iii) Compare the given equation with $$\left| {z - {z_1}} \right| + \left| {z - {z_2}} \right| = 2a$$ where $2a=10$ which is greater than $\left| {{z_1} - {z_2}} \right|$. therefore, the locus of $z$ represents the ellipse.
(iv) Compare the given equation with $$\left| {z - {z_1}} \right| - \left| {z - {z_2}} \right| = 2a$$ where $2a=10$ which is less than $\left| {{z_1} - {z_2}} \right|$. therefore, the locus of $z$ represents the hyperbola.
