PROPERTIES OF CONJUGATE - PUNJAB-CURRICULUM-AND-TEXTBOOK-BOARD-PCTB / 10TH-CLASS / MATHEMATICS / MATHEMATICS-10-10TH-CLASS-NBF-2024 / COMPLEX-NUMBERS / PROPERTIES-OF-CONJUGATE

Example 12

If $z_{1} = 3 + 4i, z_{2} = 2 + 3i$ , then prove that $\overline{z_{1} z_{2}} = \overline{z_{1}} \overline{z_{2}}$ .

Solution

$\overline{z_{1} z_{2}} = (3 + 4 i) (2 + 3 i)$

$= 6 + 9i + 8i + 12i^{2}$

$= 6 + 17i - 12$

$= -6 + 17i$

$\overline{z_{1} z_{2}} = - 6 - 17 i$

and $\overline{z_{1} z_{2}} = (3 - 4 i) (2 - 3 i)$

$= 6 - 9i - 8i + 12i^{2}$

$= 6 - 9i - 8i - 12$

$\overline{z_{1} z_{2}} = - 6 - 17 i$

From (i) and (ii), we get $\overline{z_{1} z_{2}} = \overline{z_{1}} \overline{z_{2}}$

Property 3

$(\frac{z_{1}}{z_{2}}) = \frac{\overline{z_{1}}}{\overline{z_{2}}}$ , $(z_{2}\neq 0)$

Proof: Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$

Now, $\frac{z_{1}}{z_{2}} = \frac{x_{1} + i y_{1}}{x_{2} + i y_{2}} = \frac{x_{1} + i y_{1}}{x_{2} + i y_{2}} \times \frac{x_{2} - i y_{2}}{x_{2} - i y_{2}} = \frac{(x_{1} x_{2} + y_{1} y_{2}) - i (x_{1} y_{2} - y_{1} x_{2})}{x_{2}^{2} + y_{2}^{2}}$

and $(\frac{z_{1}}{z_{2}}) = \frac{(x_{1} x_{2} + y_{1} y_{2}) + i (x_{1} y_{2} - y_{1} x_{2})}{x_{2}^{2} + y_{2}^{2}}$ ... (i)

Now, $\frac{\overline{z_{1}}}{\overline{z_{2}}} = \frac{x_{1} - i y_{1}}{x_{2} - i y_{2}}$

$= \frac{(x_{1} - i y_{1}) (x_{2} + i y_{2})}{(x_{2} - i y_{2}) (x_{2} + i y_{2})}$

$= \frac{x_{1} x_{2} + i x_{1} y_{2} - i y_{1} x_{2} - i^{2} y_{1} y_{2}}{x_{2}^{2} + y_{2}^{2}}$

$\frac{\overline{z_{1}}}{\overline{z_{2}}} = \frac{(x_{1} x_{2} + y_{1} y_{2}) + i (x_{1} y_{2} - y_{1} x_{2})}{x_{2}^{2} + y_{2}^{2}}$ ... (ii)

From (i) and (ii), we get $(\frac{z_{1}}{z_{2}}) = \frac{\overline{z_{1}}}{\overline{z_{2}}}$

Example 13

If $z_{1} = 5 + 4i, z_{2} = 3 + 2i$ , then prove that $(\frac{z_{1}}{z_{2}}) = \frac{\overline{z_{1}}}{\overline{z_{2}}}$ .

Solution

From (i) and (ii), we get

1.3 Properties of Conjugate

Example 12

Solution

Property 3

Example 13

Solution