Math, asked by sajan6491, 10 days ago

 \rm \color{blue}If  \:  {p}^{2}  +  {q}^{2}  +  {r}^{2}  = 1 \: and \: q + ir = (1 + p)z \\   \color{green} \rm then \: prove \: that \\  \rm \red{ \frac{1 - iz}{1 + iz} =  \frac{p + iq}{1 + r}  }

Answers

Answered by mathdude500
2

Appropriate Question :-

\rm \color{blue}If \: {p}^{2} + {q}^{2} + {r}^{2} = 1 \: and \: q + ir = (1 + p)z \\ \color{green} \rm then \: prove \: that \\ \rm \red{ \frac{1 - iz}{1 + iz} = \frac{p  -  iq}{1  -  r} } \\

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

Given that,

\rm \: q + ir = (1 + p)z \\

On multiply both sides by i, we get

\rm \: iq +  {i}^{2} r = (1 + p)iz \\

\rm \: iq  -  r = (1 + p)iz \:  \:  \:  \{ \because \:  {i}^{2}  =  - 1 \}   \\

\rm\implies \:\dfrac{1}{iz}  = \dfrac{p + 1}{iq - r}  \\

On applying Componendo and Dividendo, we get

\rm \:\dfrac{1 + iz}{1 - iz}  = \dfrac{p + 1 + iq - r}{p + 1 - iq  +  r}  \\

can be further rewritten as

\rm \: \dfrac{1 - iz}{1 + iz}  = \dfrac{p + 1 + r - iq}{p + 1 - r + iq}  \\

\rm \: \dfrac{1 - iz}{1 + iz}  = \dfrac{(p + 1 + r) - iq}{(p + 1 - r) + iq}  \\

On rationalizing the denominator on RHS, we get

\rm \:  = \dfrac{(p + 1 + r) - iq}{(p + 1 - r) + iq} \times \dfrac{(p + 1 - r) - iq}{(p + 1 - r) - iq}   \\

\rm \:  = \dfrac{[ {(p + 1)}^{2} -  {r}^{2}] - iq(p + 1 + r + p + 1 - r) + {i}^{2} {q}^{2} }{ {(p + 1 - r)}^{2} -  {i}^{2} {q}^{2} }  \\

\rm \:  = \dfrac{( {p}^{2} + 1 + 2p -  {r}^{2}) - iq(2p + 2) -  {q}^{2}  }{ {p}^{2}  + 1 +  {r}^{2} + 2p - 2r - 2rp +  {q}^{2}  }

\rm \:  = \dfrac{{p}^{2} + 1 + 2p -  ({r}^{2} +  {q}^{2} ) - iq(2p + 2)}{({p}^{2}  +  {q}^{2}+  {r}^{2}) + 1 + 2p - 2r - 2rp}

\rm \:  = \dfrac{{p}^{2} + 1 + 2p -  (1 - {p}^{2} ) - iq(2p + 2)}{1 + 1 + 2p - 2r - 2rp}

\rm \:  = \dfrac{{p}^{2} + 1 + 2p -  1  + {p}^{2} - 2iq(p + 1)}{1 + 1 + 2p - 2r - 2rp}

\rm \:  = \dfrac{2{p}^{2} + 2p  - 2iq(p + 1)}{2 + 2p - 2r - 2rp}

\rm \:  = \dfrac{2p(p + 1)  - 2iq(p + 1)}{2(1 +p) - 2r(1 + p)}

\rm \:  = \dfrac{2(p + 1) \: [p  - iq]}{2(1 +p) \: [1- r]}

\rm \:  = \dfrac{p - iq}{1 - r}  \\

Hence,

\color{green}\rm\implies \: \dfrac{1 - iz}{1 + iz}   = \dfrac{p - iq}{1 - r}  \\

\rule{190pt}{2pt}

Additional Information :-

Argument of complex number :-

\color{red}\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}

Similar questions