Question

Prove z1/z2 whole bar is equal to z1 bar/z2 bar.
Bar here means conjugate

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\sf{ \: If \: z = x+iy \: \: then \: \: \overline{z} = x - iy\: }$

TO PROVE

$\displaystyle \sf{ \overline{\bigg(\frac{z_1}{z_2} \bigg) }\: = \frac{ \overline{z_1}}{\overline{z_2}} }$

PROOF

$\sf{ Let \: \: z_1 = a+ib \: \: and \: \: z_2= c+id \: }$

$\sf{ \therefore \: \: \: z_1 = a - ib \: \: and \: \: z_2= c - id \: }$

Now

$\displaystyle \sf{ \frac{z_1}{z_2} \: }$

$= \displaystyle \sf{ \frac{a + ib}{c + id} \: }$

$= \displaystyle \sf{ \frac{(a + ib)(c - id)}{(c + id)( c- id)} \: }$

$= \displaystyle \sf{ \frac{ac - iad + ibc- {i}^{2}bd }{ {c}^{2} - {i}^{2} {d}^{2} } \: }$

$= \displaystyle \sf{ \frac{(ac + bd) - i(ad - bc)}{ {c}^{2} + {d}^{2} } \: }$

LHS

$= \displaystyle \sf{ \overline{\bigg(\frac{z_1}{z_2} \bigg) }\: }$

$= \displaystyle \sf{ \frac{(ac + bd) + i(ad - bc)}{ {c}^{2} + {d}^{2} } \: }$

$= \displaystyle \sf{ \frac{ac + iad - ibc- {i}^{2}bd }{ {c}^{2} - {i}^{2} {d}^{2} } \: }$

$= \displaystyle \sf{ \frac{(a - ib)(c + id)}{(c + id)( c- id)} \: }$

$= \displaystyle \sf{ \frac{(a - ib)}{( c- id)} \: }$

$\displaystyle \sf{ = \frac{ \overline{z_1}}{\overline{z_2}} }$

= RHS

Hence proved

━━━━━━━━━━━━━━━━

LEARN MORE FROM BRAINLY

show that sinix=isinhx

https://brainly.in/question/11810908

Answer 2

Answer:

adnan ayub bs Mathematics book seventh edition by James ward and sterwat