Question

Find the square root of $2x+(x^2-1)i$​

Answer 1

Basic Identities Used :-

$\boxed{ \sf \: {i}^{2} = 1}$

$\boxed{ \sf \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}}$

$\boxed{ \sf \: {(x - y)}^{2} + 4xy = {(x + y)}^{2}}$

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

$\green{\sf \: Let \: \sqrt{2x + {(x}^{2} - 1)i } = a + ib }- - (1)$

On squaring both sides, we get

$\rm :\longmapsto\:2x +({x}^{2} - 1)i = {a}^{2} + {i}^{2} {b}^{2} + 2abi$

$\rm :\longmapsto\:2x +({x}^{2} - 1)i = {a}^{2} - {b}^{2} + 2abi$

On comparing, we get

$\red{\rm :\longmapsto\: {a}^{2} - {b}^{2} = 2x} - - - (2) \\ \red{\rm :\longmapsto\:2ab = {x}^{2} - 1} - - - (3)$

Now,

Consider,

$\rm :\longmapsto\: {a}^{2}+{b}^{2} = \sqrt{ {( {a}^{2} - {b}^{2})}^{2} + {(2ab)}^{2} }$

$\rm :\longmapsto\: {a}^{2}+{b}^{2} = \sqrt{ {(2x)}^{2} + {( {x}^{2} - 1) }^{2} }$

$\rm :\longmapsto\: {a}^{2}+{b}^{2} = \sqrt{ {4x}^{2} + {( {x}^{2} - 1) }^{2} }$

$\rm :\longmapsto\: {a}^{2}+{b}^{2} = \sqrt{{( {x}^{2} + 1) }^{2} }$

$\red{\rm :\implies\:\: {a}^{2}+{b}^{2} = {x}^{2} + 1} - - - (4)$

On adding equation (2) and equation (4), we get

$\red{\rm :\implies\:\: 2{a}^{2} = {x}^{2} + 1 + 2x}$

$\red{\rm :\implies\:\: 2{a}^{2} = {(x + 1)}^{2}}$

$\red{\rm :\implies\:\: {a}^{2} = \dfrac{{(x + 1)}^{2}}{2} }$

$\red{\rm :\implies\:\: {a} = \pm \: \dfrac{{(x + 1)}}{ \sqrt{2}}} - - - (5)$

On Subtracting equation (2) from equation (4), we get

$\red{\rm :\implies\:\: 2{b}^{2} = {x}^{2} + 1 - 2x}$

$\red{\rm :\implies\:\: 2{b}^{2} = {(x - 1)}^{2}}$

$\red{\rm :\implies\:\: {b}^{2} = \dfrac{{(x - 1)}^{2}}{2} }$

$\red{\rm :\implies\:\: {b} = \pm \: \dfrac{{(x - 1)}}{ \sqrt{2}}} - - - (6)$

On substituting the values of a and b in equation (1), we get

$\green{\sf \:\sqrt{2x+{(x}^{2}- 1)i} = \pm \:\bigg(\dfrac{{(x + 1)}}{\sqrt{2}}+ i\dfrac{{(x - 1)}}{ \sqrt{2}} \bigg)}$