Question

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Answer 1

Given:-

• $\sf{z_1\:and\:z_2}$ are two complex numbers.

-

To prove:-

• $\sf{\overline{\bigg({\dfrac{z_1}{z_2}\bigg)}}=\dfrac{\overline{z_1}}{\overline{z_2}}}$

Solution:-

Let,

$\sf{z_1=a_1+i\;b_1}$ and

$\sf{z_2=a_2+i\;b_2}$

• So, Firstly calculating  $\sf{\dfrac{z_1}{z_2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1+i\;b_1}{a_2+i\;b_2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1+i\;b_1}{a_2+i\;b_2}\times \dfrac{a_2-i\;b_2}{a_2-i\;b_2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1a_2+i\;b_1a_2-a_1\;i\;b_2-i^2\;b_1b_2}{{a_2}^2-i^2\;{b_2}^2}}$

putting i² = -1

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1a_2+i(b_1a_2-a_1b_2)-(-1)\;b_1b_2}{{a_2}^2-(-1)\;{b_2}^2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1a_2+i(b_1a_2-a_1b_2)+b_1b_2}{{a_2}^2+{b_2}^2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\bigg(\dfrac{a_1a_2+b_1b_2}{{a_2}^2+{b_2}^2}\bigg) +i\bigg(\dfrac{a_2b_1-a_1b_2}{{a_2}^2+{b_2}^2}\bigg)}$

• Therefore, conjugate of  $\sf{\dfrac{z_1}{z_2}}$  would be

$\boxed{\longmapsto\sf{\overline{\bigg(\dfrac{z_1}{z_2}\bigg)}=\bigg(\dfrac{a_1a_2+b_1b_2}{{a_2}^2+{b_2}^2}\bigg) -i\bigg(\dfrac{a_2b_1-a_1b_2}{{a_2}^2+{b_2}^2}\bigg)\;\;\;....eqn(1)}}$

• Now, calculating  $\sf{\dfrac{\;\overline{z_1}\;}{\overline{z_2}}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=\overline{z_1}\times\dfrac{1}{\overline{z_2}}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=(a_1-i\;b_1)\times\dfrac{1}{(a_2-i\;b_2)}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=(a_1-i\;b_1)\times\dfrac{1}{(a_2-i\;b_2)}\times\dfrac{a_2+i\;b_2}{a_2+i\;b_2}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=(a_1-i\;b_1)\times\dfrac{a_2+i\;b_2}{({a_2}^2+{b_2}^2)}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=\dfrac{a_1a_2+i\;a_1b_2-1\;a_2b_1-i^2\;b_1b_2}{({a_2}^2+{b_2}^2)}}$

again putting i² = -1

we will get,

$\boxed{\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=\bigg(\dfrac{a_1a_2+b_1b_2}{{a_2}^2+{b_2}^2}\bigg)-i\bigg(\dfrac{a_2b_1-a_1b_2}{{a_2}^2+{b_2}^2}\bigg)}\;\;\;....eqn(2)}$

From equation (1) and (2)

$\bf{\overline{\bigg({\dfrac{z_1}{z_2}\bigg)}}=\dfrac{\overline{z_1}}{\overline{z_2}}}$  is proved.

Answer 2

Answer:

Given:-

$\sf{z_1\:and\:z_2}$ are two complex numbers.

-

To prove:-

$\sf{\overline{\bigg({\dfrac{z_1}{z_2}\bigg)}}=\dfrac{\overline{z_1}}{\overline{z_2}}}$

Solution:-

Let,

$\sf{z_1=a_1+i\;b_1}$ and

$\sf{z_2=a_2+i\;b_2}$

So, Firstly calculating  $\sf{\dfrac{z_1}{z_2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1+i\;b_1}{a_2+i\;b_2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1+i\;b_1}{a_2+i\;b_2}\times \dfrac{a_2-i\;b_2}{a_2-i\;b_2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1a_2+i\;b_1a_2-a_1\;i\;b_2-i^2\;b_1b_2}{{a_2}^2-i^2\;{b_2}^2}}$

putting i² = -1

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1a_2+i(b_1a_2-a_1b_2)-(-1)\;b_1b_2}{{a_2}^2-(-1)\;{b_2}^2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\dfrac{a_1a_2+i(b_1a_2-a_1b_2)+b_1b_2}{{a_2}^2+{b_2}^2}}$

$\longmapsto\sf{\dfrac{z_1}{z_2}=\bigg(\dfrac{a_1a_2+b_1b_2}{{a_2}^2+{b_2}^2}\bigg) +i\bigg(\dfrac{a_2b_1-a_1b_2}{{a_2}^2+{b_2}^2}\bigg)}$

Therefore, conjugate of  $\sf{\dfrac{z_1}{z_2}}$  would be

$\boxed{\longmapsto\sf{\overline{\bigg(\dfrac{z_1}{z_2}\bigg)}=\bigg(\dfrac{a_1a_2+b_1b_2}{{a_2}^2+{b_2}^2}\bigg) -i\bigg(\dfrac{a_2b_1-a_1b_2}{{a_2}^2+{b_2}^2}\bigg)\;\;\;....eqn(1)}}$

Now, calculating  $\sf{\dfrac{\;\overline{z_1}\;}{\overline{z_2}}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=\overline{z_1}\times\dfrac{1}{\overline{z_2}}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=(a_1-i\;b_1)\times\dfrac{1}{(a_2-i\;b_2)}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=(a_1-i\;b_1)\times\dfrac{1}{(a_2-i\;b_2)}\times\dfrac{a_2+i\;b_2}{a_2+i\;b_2}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=(a_1-i\;b_1)\times\dfrac{a_2+i\;b_2}{({a_2}^2+{b_2}^2)}}$

$\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=\dfrac{a_1a_2+i\;a_1b_2-1\;a_2b_1-i^2\;b_1b_2}{({a_2}^2+{b_2}^2)}}$

again putting i² = -1

we will get,

$\boxed{\longmapsto\sf{\dfrac{\;\overline{z_1}\;}{\;\overline{z_2}\;}=\bigg(\dfrac{a_1a_2+b_1b_2}{{a_2}^2+{b_2}^2}\bigg)-i\bigg(\dfrac{a_2b_1-a_1b_2}{{a_2}^2+{b_2}^2}\bigg)}\;\;\;....eqn(2)}$

From equation (1) and (2)

$\bf{\overline{\bigg({\dfrac{z_1}{z_2}\bigg)}}=\dfrac{\overline{z_1}}{\overline{z_2}}}$  is proved.