Question

Represent in polar form $\frac{1 + i}{1 - i}$

Answer 1

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

Given complex number is

$\rm :\longmapsto\:\dfrac{1 + i}{1 - i}$

Let we reduce this complex number to standard form by using Method of Rationalization.

So,

$\rm \:  =  \: \dfrac{1 + i}{1 - i} \times \dfrac{1 + i}{1 + i}$

$\rm \:  =  \: \dfrac{ {(1 + i)}^{2} }{ {1}^{2} - {i}^{2} }$

$\rm \:  =  \: \dfrac{1 + {i}^{2} + 2i}{1 - ( - 1)}$

$\rm \:  =  \: \dfrac{1 - 1 + 2i}{1 - ( - 1)}$

$\rm \:  =  \: \dfrac{ 2i}{2}$

$\rm \:  =  \: i$

$\rm \implies\:\dfrac{1 + i}{1 - i} = i$

Now, we reduce this complex number to Polar form.

Let assume that

$\rm \implies\:\dfrac{1 + i}{1 - i} = i = r(cos \theta + i \: sin \theta ) - - - (1)$

can be rewritten as

$\rm :\longmapsto\:i = rcos \theta + i \: rsin \theta$

So, on comparing Real and Imaginary parts, we get

$\red{\rm :\longmapsto\:rcos \theta = 0 - - - (2)}$

and

$\red{\rm :\longmapsto\:rsin \theta = 1 - - (3)}$

Now, Squaring equation (2) and (3) and adding, we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2} \theta + {r}^{2} {sin}^{2} \theta = 0 + 1$

$\rm :\longmapsto\: {r}^{2} ({cos}^{2} \theta + {sin}^{2} \theta ) = 1$

$\rm :\longmapsto\: {r}^{2} = 1$

$\bf\implies \:r = 1$

On substituting the value of r in equation (2) and (3), we get

$\rm :\longmapsto\:cos \theta = 0 \: \: and \: \: sin \theta = 1$

$\bf\implies \: \theta = \dfrac{\pi}{2}$

So, Equation (1) can be rewritten as

$\rm :\longmapsto\:\dfrac{1 + i}{1 - i} = cos\bigg[\dfrac{\pi}{2} \bigg] + i \: sin\bigg[\dfrac{\pi}{2} \bigg]$

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Additional Information :-

Argument of complex number

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