Maths > Complex Numbers > 5.0 Representation of complex number
Complex Numbers
1.0 Definition
2.0 Algebraic operations
3.0 Conjugate of complex number
4.0 Properties of conjugate
5.0 Representation of complex number
5.1 Cartesian form (Geometric Representation)
5.2 Principal value of complex number
5.3 Properties of modulus
5.4 Trigonometric/Polar Representation
6.0 Euler's formulae
7.0 Properties of Argument
8.0 De Moivre's Theorem
9.0 Square root of a complex number
10.0 The ${n^{th}}$ root of unity
11.0 Cube roots of unity
12.0 Rotation
13.0 Geometrical properties
14.0 Locus
15.0 Ptolemy's Theorem
5.3 Properties of modulus
5.2 Principal value of complex number
5.3 Properties of modulus
5.4 Trigonometric/Polar Representation
1. $\left| z \right| = 0 \Rightarrow z = 0$
2. $\left| z \right| = \left| {\overline z } \right| = \left| { - \overline z } \right| = \left| { - \overline z } \right|$
3. $z\overline z = {\left| z \right|^2}$
4. $\left| {{z_1}{z_2}} \right| = \left| {{z_1}} \right|\left| {{z_2}} \right|$
5. $\left| {\frac{{{z_1}}}{{{z_2}}}} \right| = \frac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}$
6. $\left| {{z_1} \pm {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right|$
7. $\left| {{z_1} \pm {z_2}} \right| \geqslant \left| {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right|$
8. $\left| {{z^n}} \right| = {\left| z \right|^n}$
9. $\left| {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right| \leqslant \left| {{z_1} \pm {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right|$
Proof: Using triangle inequality in $\Delta OAC$, as shown in figure
$$\begin{equation} \begin{aligned} OC \leqslant OA + AC \\ OA \leqslant AC + OC \\ AC \leqslant OA + OC \\\end{aligned} \end{equation} $$ Using these inequalities, we have $\left| {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right| \leqslant \left| {{z_1} + {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right|$
and
Similarly from $\Delta OAB$, we have $\left| {\left| {{z_1}} \right| - \left| {{z_2}} \right|} \right| \leqslant \left| {{z_1} - {z_2}} \right| \leqslant \left| {{z_1}} \right| + \left| {{z_2}} \right|$
10. ${\left| {{z_1} \pm {z_2}} \right|^2} = \left| {{z_1} \pm {z_2}} \right|\overline {\left| {{z_1} \pm {z_2}} \right|} = \left| {{z_1} \pm {z_2}} \right|\left| {\overline {{z_1}} \pm \overline {{z_2}} } \right|$
${\text{ = }}{\left| {{z_1}} \right|^2} + {\left| {{z_2}} \right|^2} \pm \left( {{z_1}\overline {{z_2}} + \overline {{z_1}} {z_2}} \right)$
Question 4. If $\left| {{z_1}} \right| = 1,{\text{ }}\left| {{z_2}} \right| = 1,{\text{ }}\left| {{z_3}} \right| = 1,{\text{ }}\left| {{z_1} + {z_2} + {z_3}} \right| = k$, find the value of $$\left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}}} \right|$$
Solution: As we know that, $z\overline z = {\left| z \right|^2}$, therefore,
$$\begin{equation} \begin{aligned} {z_1}\overline {{z_1}} = 1,{\text{ }}{z_2}\overline {{z_2}} = 1,{\text{ }}{z_3}\overline {{z_3}} = 1 \\ \Rightarrow \frac{1}{{{z_1}}} = \overline {{z_1}} ,{\text{ }}\frac{1}{{{z_2}}} = \overline {{z_2}} ,{\text{ }}\frac{1}{{{z_3}}} = \overline {{z_3}} \\ \therefore \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}}} \right| = \left| {\overline {{z_1}} + \overline {{z_2}} + \overline {{z_3}} } \right| = \left| {\overline {{z_1} + {z_2} + {z_3}} } \right| \\ \because \left| z \right| = \overline {\left| z \right|} \\ \Rightarrow \left| {\frac{1}{{{z_1}}} + \frac{1}{{{z_2}}} + \frac{1}{{{z_3}}}} \right| = \left| {{z_1} + {z_2} + {z_3}} \right| = k \\\end{aligned} \end{equation} $$
Question 5. If $\left| {\frac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}} \right| = 1$ in which $\left| {{z_2}} \right| \ne 1$. Find $\left| {{z_1}} \right|$.
Solution: $$\begin{equation} \begin{aligned} \left| {\frac{{{z_1} - 2{z_2}}}{{2 - {z_1}\overline {{z_2}} }}} \right| = 1 \\ \left| {{z_1} - 2{z_2}} \right| = \left| {2 - {z_1}\overline {{z_2}} } \right| \\\end{aligned} \end{equation} $$
Squaring both sides, we get
$$\begin{equation} \begin{aligned} {\left| {{z_1} - 2{z_2}} \right|^2} = {\left| {2 - {z_1}\overline {{z_2}} } \right|^2} \\ \left| {{z_1} - 2{z_2}} \right|\overline {\left| {{z_1} - 2{z_2}} \right|} = \left| {2 - {z_1}\overline {{z_2}} } \right|\overline {\left| {2 - {z_1}\overline {{z_2}} } \right|} \\ \left( {{z_1} - 2{z_2}} \right)\left( {\overline {{z_1}} - 2\overline {{z_2}} } \right) = \left( {2 - {z_1}\overline {{z_2}} } \right)\left( {2 - \overline {{z_1}} {z_2}} \right) \\ {\left| {{z_1}} \right|^2} - 2{z_1}\overline {{z_2}} - 2{z_1}\overline {{z_2}} + 4{\left| {{z_2}} \right|^2} = 4 - 2\overline {{z_1}} {z_2} - 2\overline {{z_1}} {z_2} + {\left| {{z_1}} \right|^2}{\left| {{z_2}} \right|^2} \\ \left( {{{\left| {{z_1}} \right|}^2} - 4} \right) - {\left| {{z_2}} \right|^2}\left( {{{\left| {{z_1}} \right|}^2} - 4} \right) = 0 \\ \left( {{{\left| {{z_1}} \right|}^2} - 4} \right)\left( {1 - {{\left| {{z_2}} \right|}^2}} \right) = 0 \\ {\left| {{z_1}} \right|^2} = 4 \Rightarrow {z_1} = 2 \\ {\left| {{z_2}} \right|^2} = 1 \Rightarrow {z_2} = 1,{\text{ not possible}}{\text{.}} \\\end{aligned} \end{equation} $$
Question 6. Find the greatest and least value of the modulus of complex number $z$ satisfying the equation $$\left| {z - \frac{4}{z}} \right| = 2$$
Solution: From figure, we have 
$$\begin{equation} \begin{aligned} \left| {OP - OQ} \right| \leqslant QP \\ \Rightarrow \left| {\left| z \right| - \left| {\frac{4}{z}} \right|} \right| \leqslant \left| {z - \frac{4}{z}} \right| = 2 \\ \Rightarrow - 2 \leqslant \left| z \right| - \left| {\frac{4}{z}} \right| \leqslant 2 \\ \Rightarrow {\left| z \right|^2} + 2\left| z \right| - 4 \geqslant 0{\text{ or }}{\left| z \right|^2} - 2\left| z \right| - 4 \leqslant 0 \\ {\left( {\left| z \right| + 1} \right)^2} - 5 \geqslant 0{\text{ or }}{\left( {\left| z \right| - 1} \right)^2} - 5 \leqslant 0 \\ \left( {\left| z \right| + 1 + \sqrt 5 } \right)\left( {\left| z \right| + 1 - \sqrt 5 } \right) \geqslant 0{\text{ or }}\left( {\left| z \right| - 1 + \sqrt 5 } \right)\left( {\left| z \right| - 1 - \sqrt 5 } \right) \leqslant 0 \\ \Rightarrow \sqrt 5 - 1 \leqslant \left| z \right| \leqslant \sqrt 5 + 1 \\\end{aligned} \end{equation} $$
Therefore, greatest value of ${\left| z \right|}$ is $\sqrt 5 + 1$ and least value of ${\left| z \right|}$ is $\sqrt 5 - 1$.
