Question

find the value of
$\sqrt{i}$
​

Answer 1

see the image attached by me ✔️✔️

(what happened to u friend) ☹️

Answer 2

Question:

Find the value of $\sqrt{i}$.

Answer:

$The\: value \: of \: \underline{\sf\pink{\sqrt{i \: }} \: is \: \sf\pink{\frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } i}}$

Step-by-step explanation:

$\sqrt{i \: } = \: ?$

Where, ${(i)}^{2} = - 1$

Let us start with complex number (i.e) $x\:or\:y\:or\:z$

$z = x + iy$

$\therefore {z}^{2} = i$

$\longmapsto {z}^{2} = {(x + iy)}^{2}$

$\longmapsto {z}^{2} = {x}^{2} + 2xiy + {i}^{2} {y}^{2}$

$\longmapsto {z}^{2} = ({x}^{2} - {y}^{2} ) + 2xiy$

We know that,

$\boxed{ {z}^{2} = i = 0 +1 i} \: thus,$

$\bull \: ( {x}^{2} - {y}^{2} ) = 0 \: ---> (1)$

$\bull \: 2xy = 1 \: ---> (2)$

Now, ( 1 )

$\hookrightarrow {x}^{2} = {y}^{2}$

$\hookrightarrow \boxed{ y = ±x}$

Now substituting the value of y in ( 2 )

$\rightsquigarrow 2 {x}^{2} = ±x$

$\rightsquigarrow \boxed{{x}^{2} = ± \frac{1}{2} }$

$\therefore$ The values are,

$\bull \: x = y = \frac{1}{ \sqrt{2} }$

$\bull \: x = y = - \frac{1}{ \sqrt{2} }$

$\bull \: x =\frac{1}{ \sqrt{2} } i, \: y = - \frac{1}{ \sqrt{2} } i$

$\bull \: x = - \frac{1}{ \sqrt{2} } i, \: y = \frac{1}{ \sqrt{2} } i$

Now, ( a )

$\Longrightarrow \: z = \frac{1}{ \sqrt{2} } i \: - {\big(\: \frac{1}{ \sqrt{2} } i\big)}^{2}$

$\Longrightarrow \: \underline{ \: z = \frac{1}{ \sqrt{2} } \: + \: \frac{1}{ \sqrt{2} } i \: }$

Now, ( b )

$\longrightarrow \: z = - \frac{1}{ \sqrt{2} } i \: + {\big(\: \frac{1}{ \sqrt{2} } i\big)}^{2}$

$\longrightarrow \: \underline{ \: z = - \frac{1}{ \sqrt{2} } \: - \: \frac{1}{ \sqrt{2} } i \: }$

Thus, $z = \frac{1}{ \sqrt{2} } \: + \: \frac{1}{ \sqrt{2} } i\:and\:- \frac{1}{ \sqrt{2} } \: - \: \frac{1}{ \sqrt{2} } i$

Now evaluating principle $\underline{ \ :\sqrt{i} \:}$

$\therefore \: \boxed { \bf \sqrt{i \: } = \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } i}$