Question

Solve both the questions in the attachment!!

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Answer 1

$\large\underline{\sf{Solution-22}}$

Consider,

$\rm :\longmapsto\:\dfrac{ \sqrt{7} + i \sqrt{3} }{ \sqrt{7} - i \sqrt{3}} + \dfrac{ \sqrt{7} - i \sqrt{3} }{ \sqrt{7} + i \sqrt{3} }$

On taking LCM, we get

$\rm \:  =  \: \dfrac{ {( \sqrt{7} + i \sqrt{3})}^{2} + {( \sqrt{7} - i \sqrt{3})}^{2} }{( \sqrt{7} + i \sqrt{3})( \sqrt{7} - i \sqrt{3})}$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}}$

So, using these Identities, we get

$\rm \:  =  \: \dfrac{2\bigg[ {( \sqrt{7})}^{2} + {(i \sqrt{3}) }^{2} \bigg]}{ {( \sqrt{7}) }^{2} - {(i \sqrt{3})}^{2} }$

$\rm \:  =  \: \dfrac{2(7 + 3 {i}^{2})}{7 - {3i}^{2} }$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {i}^{2} = - 1}}}$

So, using this, we get

$\rm \:  =  \: \dfrac{2(7 - 3)}{7 + 3}$

$\rm \:  =  \: \dfrac{2 \times 4}{10}$

$\rm \:  =  \: \dfrac{4}{5}$

Hence,

$\red{\rm\implies \:\boxed{\sf{ \dfrac{ \sqrt{7} + i \sqrt{3} }{ \sqrt{7} - i \sqrt{3}} + \dfrac{ \sqrt{7} - i \sqrt{3} }{ \sqrt{7} + i \sqrt{3} } \: is \: purely \: real}}}$

$\green{\large\underline{\sf{Solution-23}}}$

Given that,

$\rm :\longmapsto\: {(x + iy)}^{3} = y + iv$

$\rm :\longmapsto\: {x}^{3} + 3 {x}^{2}(iy) + 3x {(iy)}^{2} + {(iy)}^{3} = y + iv$

$\rm :\longmapsto\: {x}^{3} + 3 {x}^{2}yi + 3x {i}^{2} {y}^{2} + {i}^{3} {y}^{3} = y + iv$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {i}^{2} = - 1 \: \: \: and \: \: \: {i}^{3} = - i }}}$

So, using this, we get

$\rm :\longmapsto\: {x}^{3} + 3 {x}^{2}yi - 3x{y}^{2} - i{y}^{3} = y + iv$

$\rm :\longmapsto\: ({x}^{3} - 3x {y}^{2}) + i(3{x}^{2}y- {y}^{3})= y + iv$

On comparing real and Imaginary parts, we get

$\rm :\longmapsto\:y = {x}^{3} - 3 {xy}^{2}$

and

$\rm :\longmapsto\:v = {3x}^{2}y - {y}^{3}$

Now, Consider

$\rm :\longmapsto\:\dfrac{y}{x} + \dfrac{v}{y}$

$\rm \:  =  \: \dfrac{ {x}^{3} - {3xy}^{2}}{x} + \dfrac{ {3yx}^{2} - {y}^{3} }{y}$

$\rm \:  =  \: \dfrac{ x({x}^{2} - {3y}^{2})}{x} + \dfrac{y({3x}^{2} - {y}^{2})}{y}$

$\rm \:  =  \: {x}^{2} - {3y}^{2} + {3x}^{2} - {y}^{2}$

$\rm \:  =  \: 4{x}^{2} - {4y}^{2}$

$\rm \:  =  \: 4({x}^{2} - {y}^{2})$

Hence, Proved

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MORE TO KNOW

Argument of complex number

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Complex \: number & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf  {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg|   \\ \\ \sf  - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf  - x - iy & \sf  - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf  - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}} \\ \end{gathered}$

Answer 2

$\large\underline{\sf{Solution-22}} \\ \\ Consider, \\ \\ \rm :\longmapsto\:\dfrac{ \sqrt{7} + i \sqrt{3} }{ \sqrt{7} - i \sqrt{3}} + \dfrac{ \sqrt{7} - i \sqrt{3} }{ \sqrt{7} + i \sqrt{3} } \\ \\ On \: \: taking \: \: LCM, \: \: we \: \: get \\ \\ \rm \: = \: \dfrac{ {( \sqrt{7} + i \sqrt{3})}^{2} + {( \sqrt{7} - i \sqrt{3})}^{2} }{( \sqrt{7} + i \sqrt{3})( \sqrt{7} - i \sqrt{3})} \\ \\ We \: \: know,\\ \\ \begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}}} \\ \end{gathered} \\ \\ and \\ \\ \begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ (x + y)(x - y) = {x}^{2} - {y}^{2} \: }}} \\ \end{gathered} \\ \\ So \: \: , using \: \: these \: \: Identities \: \: , we \: \: get \\ \\ \rm \: = \: \dfrac{2\bigg[ {( \sqrt{7})}^{2} + {(i \sqrt{3}) }^{2} \bigg]}{ {( \sqrt{7}) }^{2} - {(i \sqrt{3})}^{2} } \\ \\ \rm \: = \: \dfrac{2(7 + 3 {i}^{2})}{7 - {3i}^{2} } \\ \\ We \: \: know, \\ \\\begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ {i}^{2} = - 1}}} \\ \end{gathered} \\ \\ So, \: \: using \: \: this, \: \: we \: \: get \\ \\ \rm \: = \: \dfrac{2(7 - 3)}{7 + 3} \\ \\ \rm \: = \: \dfrac{2 \times 4}{10} \\ \\ \rm \: = \: \dfrac{4}{5} \\ \\ Hence \\ \\ \red{\rm\implies \:\boxed{\sf{ \dfrac{ \sqrt{7} + i \sqrt{3} }{ \sqrt{7} - i \sqrt{3}} + \dfrac{ \sqrt{7} - i \sqrt{3} }{ \sqrt{7} + i \sqrt{3} } \: is \: purely \: real}}}</p><p>$

$\green{\large\underline{\sf{Solution-23}}} \\ \\ Given \: \: that,</p><p> \\ \\ \rm :\longmapsto\: {(x + iy)}^{3} = y + iv:⟼(x+iy) \\ \\ \rm :\longmapsto\: {x}^{3} + 3 {x}^{2}(iy) + 3x {(iy)}^{2} + {(iy)}^{3}</p><p> \\ \\ \rm :\longmapsto\: {x}^{3} + 3 {x}^{2}yi + 3x {i}^{2} {y}^{2} + {i}^{3} {y}^{3} \\ \\ We \: \: know, \\ \\ \begin{gathered} \purple{\rm :\longmapsto\:\boxed{\tt{ {i}^{2} = - 1 \: \: \: and \: \: \: {i}^{3} = - i }}} \\ \end{gathered} \\ \\ So, \: \: using \: \: this, \: \: we \: \: get \\ \\\rm :\longmapsto\: {x}^{3} + 3 {x}^{2}yi - 3x{y}^{2} - i{y}^{3} \\ \\ \rm :\longmapsto\: ({x}^{3} - 3x {y}^{2}) + i(3{x}^{2}y- {y}^{3}) \\ \\ On \: \: comparing \: \: real \: \: and \: \: Imaginary \: \: parts, \: \: we \: \: get \\ \\ \rm :\longmapsto\:y = {x}^{3} - 3 {xy}^{2} \\ \\ and \\ \\ \rm :\longmapsto\:v = {3x}^{2}y - {y}^{3} \\ \\ Now, \: \: Consider \\ \\ \rm :\longmapsto\:\dfrac{y}{x} + \dfrac{v}{y} \\ \\ \rm \: = \: \dfrac{ {x}^{3} - {3xy}^{2}}{x} + \dfrac{ {3yx}^{2} - {y}^{3} }{y} \\ \\ \rm \: = \: \dfrac{ x({x}^{2} - {3y}^{2})}{x} + \dfrac{y({3x}^{2} - {y}^{2})}{y} \\ \\ \rm \: = \: {x}^{2} - {3y}^{2} + {3x}^{2} - {y}^{2} </p><p> \\ \\ \rm \: = \: 4{x}^{2} - {4y}^{2} \\ \\ \rm \: = \: 4({x}^{2} - {y}^{2}) \\ \\ Hence, \: \: Proved.$

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