Question

Z =1/1+i
Express in polax and Exponential form​

Answer 1

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

Given complex number is

$\rm :\longmapsto\:z = \dfrac{1}{1 + i}$

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

$\rm :\longmapsto\:z = \dfrac{1 - i}{ {1}^{2} - {i}^{2} }$

$\rm :\longmapsto\:z = \dfrac{1 - i}{1 + 1} \: \: \: \: \{ \: {i}^{2} = - 1 \: \}$

$\rm :\longmapsto\:z = \dfrac{1 - i}{2}$

$\bf\implies \:z = \dfrac{1}{2} \: - \: \dfrac{1}{2} i$

Let assume that

$\rm :\longmapsto\:\dfrac{1}{2} - \dfrac{1}{2} i = r(cosx + isinx) - - - (1)$

$\rm :\longmapsto\:\dfrac{1}{2} - \dfrac{1}{2} i = rcosx + irsinx$

On comparing real and Imaginary parts we get

$\rm :\longmapsto\:rcosx = \dfrac{1}{2} - - - (2)$

and

$\rm :\longmapsto\:rsinx = - \dfrac{1}{2} - - - (3)$

Squaring equation (2) and (3) and add, we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}x + {r}^{2} {sin}^{2}x = \dfrac{1}{4} + \dfrac{1}{4}$

$\rm :\longmapsto\: {r}^{2} ({cos}^{2}x + {sin}^{2}x) = \dfrac{2}{4}$

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

$\bf\implies \:r = \dfrac{1}{ \sqrt{2} } - - - (4)$

On substituting (4) in equation (2) and (3), we get

$\rm :\longmapsto\:cosx = \dfrac{1}{ \sqrt{2} } \: \: and \: \: sinx = - \: \dfrac{1}{ \sqrt{2} }$

We know, sinx < 0 and cosx > 0 in fourth quadrant.

$\bf\implies \:x = - \: \dfrac{\pi}{4}$

So, we have

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{r = \dfrac{1}{ \sqrt{2} } } \\ \\ &\sf{x = - \: \dfrac{\pi}{4} } \end{cases}\end{gathered}\end{gathered}$

Hence, Equation (1) become

$\red{\rm :\longmapsto\:\dfrac{1}{2} - \dfrac{1}{2} i = \dfrac{1}{ \sqrt{2} } \bigg(cos\bigg[ - \dfrac{\pi}{4} \bigg] + isin\bigg[ - \dfrac{\pi}{4} \bigg]\bigg)}$

is the required polar form.

We know,

Exponential form is

$\rm :\longmapsto\:\boxed{ \tt{ \: r(cosx + i \: sinx) = r {e}^{ix} \: }}$

So, Exponential form is

$\red{\rm :\longmapsto\:\dfrac{1}{2} - \dfrac{1}{2} i = \dfrac{1}{ \sqrt{2} } {\bigg(e\bigg)}^{ -i \dfrac{\pi}{4} } }$

Hence,

$\rm :\longmapsto\:z = \dfrac{1}{1 + i} = \dfrac{1}{ \sqrt{2} } \bigg(cos\bigg[ - \dfrac{\pi}{4} \bigg] + isin\bigg[ - \dfrac{\pi}{4} \bigg]\bigg)$

and

$\rm :\longmapsto\:z = \dfrac{1}{1 + i} = \dfrac{1}{ \sqrt{2} } {\bigg(e\bigg)}^{ -i \dfrac{\pi}{4} }$

Answer 2

Answer:

$\\ \\ </p><p></p><p>\rm :\longmapsto\:z = \dfrac{1}{1 + i} = \dfrac{1}{ \sqrt{2} } \bigg(cos\bigg[ - \dfrac{\pi}{4} \bigg] + isin\bigg[ - \dfrac{\pi}{4} \bigg]\bigg) \\ </p><p></p><p> \\ \\ and \\ \\ </p><p></p><p>\rm :\longmapsto\:z = \dfrac{1}{1 + i} = \dfrac{1}{ \sqrt{2} } {\bigg(e\bigg)}^{ -i \dfrac{\pi}{4} }</p><p></p><p>$

Step-by-step explanation:

Hope it helps you