Question

answer with explanation..​

Answer 1

Question :-

$\rm \: If \: x + iy = \dfrac{1}{1 + cos\theta + isin\theta }, \: show \: that \: {4x}^{2} - 1 = 0$

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

Given that,

$\rm \: \: x + iy = \dfrac{1}{1 + cos\theta + isin\theta }$

We know,

$\boxed{\sf{  \:\rm \: 1 + cos2x = {2cos}^{2}x \: \: }}$

and

$\boxed{\sf{  \:\rm \: sin2x = 2sinx \: cosx \: \: }}$

So, using these results, we get

$\rm \: x + iy = \dfrac{1}{2 {cos}^{2} \dfrac{\theta }{2} + 2sin\dfrac{\theta }{2}cos\dfrac{\theta }{2} }$

$\rm \: x + iy = \dfrac{1}{ 2cos\dfrac{\theta }{2}\bigg(cos\dfrac{\theta }{2} + isin\dfrac{\theta }{2}\bigg) }$

On rationalizing the denominator, we get

$\rm \: x + iy = \dfrac{1}{ 2cos\dfrac{\theta }{2}\bigg(cos\dfrac{\theta }{2} + isin\dfrac{\theta }{2}\bigg) } \times \dfrac{cos\dfrac{\theta }{2} - isin\dfrac{\theta }{2}}{cos\dfrac{\theta }{2} - isin\dfrac{\theta }{2}}$

$\rm \: x + iy = \dfrac{cos\dfrac{\theta }{2} - isin\dfrac{\theta }{2}}{ 2cos\dfrac{\theta }{2}\bigg(cos^{2} \dfrac{\theta }{2} - i^{2} sin^{2} \dfrac{\theta }{2}\bigg) }$

$\rm \: x + iy = \dfrac{cos\dfrac{\theta }{2} - isin\dfrac{\theta }{2}}{ 2cos\dfrac{\theta }{2}\bigg(cos^{2} \dfrac{\theta }{2} + sin^{2} \dfrac{\theta }{2}\bigg) } \: \: \{ \because \: {i}^{2} = - 1 \}$

$\rm \: x + iy = \dfrac{cos\dfrac{\theta }{2} - isin\dfrac{\theta }{2}}{ 2cos\dfrac{\theta }{2} } \: \: \{ \because \: {sin}^{2}x + {cos}^{2}x = 1 \}$

$\rm \: x + iy = \dfrac{1}{2} - \dfrac{1}{2} tan\dfrac{\theta }{2} \: i$

So, on comparing real part, we get

$\rm \: x = \dfrac{1}{2}$

$\rm \: 2x = 1$

$\rm \: {4x}^{2} = 1$

$\rm\implies \: {4x}^{2} - 1 = 0$

$\rule{190pt}{2pt}$

Additional Information :-

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

Step-by-step explanation:

$\color{darkcyan} \pmb{\begin{array}{l} \tt x+i y=\dfrac{1}{1+\cos \theta+i \sin \theta} \\ \\ \tt =\dfrac{1}{2 \cos ^{2} \dfrac{\theta}{2}+i 2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}} \\ \\ \tt =\dfrac{1}{2 \cos \dfrac{\theta}{2}} \times \dfrac{1}{\cos \dfrac{\theta}{2}+i \sin \dfrac{\theta}{2}} \\ \\ \tt =\dfrac{1}{2 \cos \dfrac{\theta}{2}} \times \dfrac{1}{e^{i\dfrac{\theta}{2}}} \\ \\ \tt =\dfrac{1}{2 \cos \dfrac{\theta}{2}} \times e^{-i \dfrac{\theta}{2}} \\ \\ \tt=\dfrac{1}{2 \cos \dfrac{\theta}{2}}\left(\cos \dfrac{\theta}{2}-i \sin \dfrac{\theta}{2}\right)=\dfrac{1}{2} -i \dfrac{1}{2} \tan \dfrac{\theta}{2} \\ \\ \tt \Longrightarrow x=\dfrac{1}{2} \\ \\ \tt = 4 x^{2}-1=4 \times\left(\dfrac{1}{2}\right)^{2}-1\\ \\ \tt=1-1 =0 \end{array}}$