# Complex Numbers Set 1

Go back to  'SOLVED EXAMPLES'

Example- 1

Show that if \begin{align}&\frac{1}{{{z_1} - {z_2}}} + \frac{1}{{{z_2} - {z_3}}} + \frac{1}{{{z_3} - {z_1}}} = 0,\,\,{z_1},\,\,{z_2}{\text{ and }}{z_3}\end{align}represent the vertices of an equilateral triangle.

Solution:   Let \begin{align}&{z_1},\,\,{z_2}{\text{ and }}{z_3}\end{align}  represent the vertices $$A, B$$ and $$C$$ of the triangle $$ABC.$$ We need to show that $$ABC$$ is equilateral.

We multiply the given relation by $${z_3} - {z_1}$$ on both sides to obtain:

\begin{align}&\frac{{{z_3} - {z_1}}}{{{z_1} - {z_2}}} + \frac{{{z_3} - {z_1}}}{{{z_2} - {z_3}}} + 1 = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\qquad\;\;\;\underbrace {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}_{{\rm{combine}}}\\ &\Rightarrow \,\,\,\frac{{{z_3} - {z_1}}}{{{z_1} - {z_2}}} + \frac{{{z_2} - {z_1}}}{{{z_2} - {z_3}}} = 0\\ &\Rightarrow \,\,\,\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}} = \frac{{{z_1} - {z_2}}}{{{z_3} - {z_2}}}\\ &\Rightarrow \,\,\,{\rm{Arg }}\left( {\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}}} \right) = {\rm{Arg }}\left( {\frac{{{z_1} - {z_2}}}{{{z_3} - {z_2}}}} \right)\\ &\Rightarrow \,\,\,\angle A = \angle B \end{align}

Similarly, we can prove that $$\angle B = \angle C$$  and therefore$$\angle A = \angle B = \angle C$$,. Thus, $$ABC$$ is equilateral

Example- 2

If $$\left| z \right| \le 1\,\,\,{\rm{and }}\left| w \right| \le 1,$$ show that

(i)   $${\left| {z - w} \right|^2} \le {\left( {\left| z \right| - \left| w \right|} \right)^2} + {\left( {\arg (z) - \arg (w)} \right)^2}$$

(ii)  $${\left| {z + w} \right|^2} \ge {\left( {\left| z \right| + \left| w \right|} \right)^2} - {\left( {\arg (z) - \arg (w)} \right)^2}$$

Solution:   Assume some arbitrary values for $$z$$ and $$w,$$ and plot them on the plane (both $$z$$ and w will lie inside the unit circle)

Applying the cosine rule on triangle OAB above, we obtain :

\begin{align}&A{B^2} = O{A^2} + O{B^2} - 2OA \cdot OB \cdot \cos \theta\\&{\left| {z - w} \right|^2} = {\left| z \right|^2} + {\left| w \right|^2} - 2\left| z \right|\left| w \right|\cos \theta\\&= {\left( {\left| z \right| - \left| w \right|} \right)^2} + 2\left| z \right|\left| w \right| - 2\left| z \right|\left| w \right|\cos \theta\\&= {\left( {\left| z \right| - \left| w \right|} \right)^2} + 2\left| z \right|\left| w \right|(1 - \cos \theta )\\&= {\left( {\left| z \right| - \left| w \right|} \right)^2} + 4\left| z \right|\left| w \right|{\sin ^2}\frac{\theta }{2}\end{align}

\begin{align}& \le {\left( {\left| z \right| - \left| w \right|} \right)^2} + {\theta ^2}\left( \begin{array}{l} {\rm{because}}\left| z \right| \le 1,\,\,\,\,\left| w \right| \le 1\,\,{\rm{and }}\\ \sin \frac{\theta }{2} \le \frac{\theta }{2}{\rm{ \;\;so\;\; that\;\;}}{\sin ^2}\frac{\theta }{2} \le \frac{{{\theta ^2}}}{4} \end{array} \right)\end{align}

This proves the first part.

To prove the second part, we apply the cosine rule again as shown below:

Applying the cosine rule on triangle OBC, we obtain

$\begin{array}{l}O{C^2} = O{B^2} + B{C^2} - 2OB \cdot BC \cdot \cos \,(\pi - \theta )\\ \Rightarrow \,\,\,{\left| {z + w} \right|^2} = {\left| z \right|^2} + {\left| w \right|^2} + 2\left| z \right|\left| w \right|\cos \theta \\ = {\left( {\left| z \right| + \left| w \right|} \right)^2} - 2\left| z \right|\left| w \right|(1 - \cos \theta )\\ = {\left( {\left| z \right| + \left| w \right|} \right)^2} - 4\left| z \right|\left| w \right|{\sin ^2}\frac{\theta }{2}\\ \ge {\left( {\left| z \right| + \left| w \right|} \right)^2} - {\theta ^2} & \left( \begin{array}{l}{\rm{because, \text{as in the previous}}}\\{\rm{part}},4\left| z \right|\left| w \right|{\sin ^2}\frac{\theta }{2} \le {\theta ^2}\end{array} \right)\end{array}$

Example- 3

Two different non-parallel lines meet the circle $$\left| z \right| = r$$ in the points a, b and c, d respectively. Prove that these lines meet in the point z given by

$z = \frac{{{a^{ - 1}} + {b^{ - 1}} - {c^{ - 1}} - {d^{ - 1}}}}{{{a^{ - 1}}{b^{ - 1}} - {c^{ - 1}}{d^{ - 1}}}}$

Solution: The situation given in the question is sketched in the figure below:

Since a, b, c, d lie on the circle $$\left| z \right| = r$$,  we have

$\begin{array}{l}\left| a \right| = \left| b \right| = \left| c \right| = \left| d \right| = r\\\\ \Rightarrow \,\,\,a\bar a = b\bar b = c\bar c = d\bar d = {r^2} & & \ldots (1)\end{array}$

Now, a, b and z are collinear $$\Rightarrow \,\,\,z - b = \lambda (b - a)$$

\begin{align} \Rightarrow \,\,\,\frac{{z - b}}{{b - a}} \text{ is purely real}\end{align}

\begin{align}{} \Rightarrow& \,\,\,\frac{{z - b}}{{b - a}} = \frac{{\bar z - \bar b}}{{\bar b - \bar a}}\\\\ \Rightarrow& \,\,\,(\bar b - \bar a)z - (b - a)\bar z + \bar ab - a\bar b = 0\\\\ \Rightarrow& \,\,\,\bar z = \frac{{(\bar b - \bar a)z + \bar ab - a\bar b}}{{b - a}} \ldots (2)\end{align}

Similarly, since c, d and z are collinear,

$\bar z = \frac{{(\bar d - \bar c)z + \bar cd - c\bar d}}{{d - c}} \ldots (3)$

From (2) and (3)

$\frac{{\left( {\bar b - \bar a} \right)z + \bar ab - a\bar b}}{{b - a}} = \frac{{\left( {\bar d - \bar c} \right)z + \bar cd - c\bar d}}{{d - c}} \ldots (4)$

Using (1) in (4), we obtain

\begin{align}&\frac{{\left( {\frac{{{r^2}}}{b} - \frac{{{r^2}}}{a}} \right)z + \frac{{{r^2}b}}{a} - \frac{{{r^2}a}}{b}}}{{b - a}} = \frac{{\left( {\frac{{{r^2}}}{d} - \frac{{{r^2}}}{c}} \right)z + \frac{{{r^2}d}}{c} - \frac{{{r^2}c}}{d}}}{{d - c}}\\\\ &\Rightarrow \,\,\,\frac{{ - z}}{{ab}} + \frac{{b + a}}{{ab}} = \frac{{ - z}}{{cd}} + \frac{{d + c}}{{cd}}\\\\ &\Rightarrow \,\,\,z\left( {\frac{1}{{ab}} - \frac{1}{{cd}}} \right) = \frac{{b + a}}{{ab}} - \frac{{d + c}}{{cd}}\\\\ &\qquad\qquad\qquad\qquad\;= \frac{1}{a} + \frac{1}{b} - \frac{1}{c} - \frac{1}{d}\\ &\qquad\qquad\qquad\qquad\;= {a^{ - 1}} + {b^{ - 1}} - {c^{ - 1}} - {d^{ - 1}}\\\\ &\Rightarrow \,\,\,z = \frac{{{a^{ - 1}} + {b^{ - 1}} - {c^{ - 1}} - {d^{ - 1}}}}{{{a^{ - 1}}{b^{ - 1}} - {c^{ - 1}}{d^{ - 1}}}}\end{align}

Example-  4

If $${z_1},\,\,{z_2}{\;\rm{ and }\;}{z_3}$$   represent the complex numbers A, B, C respectively and \begin{align}\angle ABC = \angle ACB = \frac{1}{2}\left( {\pi - \alpha } \right),\end{align}  prove that $${({z_2} - {z_3})^2} = 4({z_3} - {z_1})({z_1} - {z_2})\,{\sin ^2}\,\,\alpha /2$$ .

Solution:   Since $$\angle ABC = \angle ACB,$$ the triangle ABC is isosceles

Applying rotation about $${z_1}\,(\overrightarrow {AB} \to \overrightarrow {AC} )$$:

$\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}} = {e^{i\alpha }} = \cos \alpha + i\sin \alpha \;\;\;\;\; .... (1)$

Since the final result we would like to obtain contains $$\alpha /2$$, we subtract 1 from both sides of (1)

\begin{align}& \Rightarrow \,\,\,\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}} - 1 = - 1 + \cos \alpha + i\sin \alpha \\ &\Rightarrow \,\,\,\frac{{{z_3} - {z_2}}}{{{z_2} - {z_1}}} = - 2{\sin ^2}\frac{\alpha }{2} + 2i\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\\ &\qquad\qquad\quad\;\;= - 2\sin \frac{\alpha }{2}\left( {\sin \frac{\alpha }{2} - i\cos \frac{\alpha }{2}} \right)\\ &\qquad\qquad\quad\;\;= 2i\sin \frac{\alpha }{2}\left( {\cos \frac{\alpha }{2} + i\sin \frac{\alpha }{2}} \right)\\ &\qquad\qquad\quad\;\;= 2i\sin \frac{\alpha }{2}{e^{i\,\alpha /2}}\end{align}

Squaring both sides,

\begin{align}&\frac{{{{\left( {{z_3} - {z_2}} \right)}^2}}}{{{{\left( {{z_2} - {z_1}} \right)}^2}}} = - 4{\sin ^2}\frac{\alpha }{2}{e^{i\alpha }}\\ &\qquad\quad\;\;\quad= - 4{\sin ^2}\frac{\alpha }{2}\left( {\frac{{{z_3} - {z_1}}}{{{z_2} - {z_1}}}} \right)\end{align}

Cross-multiplying by $${\left( {{z_2} - {z_1}} \right)^2}$$  gives us the desired result.

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school

0