Question

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

Answer 1

SOLUTION :

❒ To Find :-

• Square root of 2x + (x² - 1)i = ?

❒ Calculation :-

Given complex number :

• 2x + (x² - 1)i

Square root of the complex number will be :

• $\sf{ \sqrt{2x + (x {}^{2} - 1)i}}$

Suppose,

• $\sf{ \sqrt{2x + (x {}^{2} - 1)i} = \alpha + i \beta}$ —————— > (1)

Squaring both side of the Eq. (1), we get,

• $\bigstar \: \: \sf{ \{ \sqrt{ 2x + (x {}^{2} - 1)i} \: \} {}^{2} =( \alpha + i \beta) {}^{2}}$

$\implies \: \sf{2x + (x {}^{2} - 1)i= \alpha {}^{2} +( i \beta) {}^{2} + 2. \alpha .i \beta }$

$\implies \: \sf{2x + i(x {}^{2} - 1)= \alpha {}^{2} + i {}^{2} \beta {}^{2} +i \: 2\alpha \beta }$

We know that,

• $\: \: \: \: \: \dag \: \: \underline{ \boxed{ \large{\rm{i {}^{2} = - 1}}}}$

Hence,

• $\star \: \: \sf{2x + i(x {}^{2} - 1)= \alpha {}^{2} + i {}^{2} \beta {}^{2} +i \: 2\alpha \beta }$

$\implies \: \sf{2x + i(x {}^{2} - 1)= \alpha {}^{2} + ( - 1)\beta {}^{2} +i \: 2 \alpha \beta }$

$\implies \: \sf{2x + i(x {}^{2} - 1)= (\alpha {}^{2} - \beta {}^{2}) +i \: (2\alpha \beta) }$

Comparing real and imaginary parts of both side, we get,

• 2x = α² - β²

• x² - 1 = 2 α β

We can consider that,

• $\bigstar \: \: \sf{\alpha {}^{2} + \beta {}^{2} = \sqrt{( \alpha {}^{2}+ \beta {}^{2}) {}^{2} }}$

$\implies \: \sf{\alpha {}^{2} + \beta {}^{2} = \sqrt{( \alpha {}^{2} + \beta {}^{2}) - 4 \alpha{}^{2} \beta {}^{2} + 4 \alpha{}^{2} \beta {}^{2}}} \\ \\ \implies \: \sf{ \alpha {}^{2} + \beta {}^{2} = \sqrt{( \alpha {}^{2} - \beta {}^{2}) {}^{2} + (2 \alpha \beta ) {}^{2}}} \: \: \: \: \: \: \: \: \: \: \:$

Substituting the values of α² - β² and 2 α β, we get,

• $\bigstar \: \: \sf{\alpha {}^{2} + \beta {}^{2} = \sqrt{( \alpha {}^{2} - \beta {}^{2}) {}^{2} + (2 \alpha \beta ) {}^{2}}}$

$\implies \: \sf{ \alpha {}^{2} + \beta {}^{2} = \sqrt{(2x) {}^{2}+ ( {x}^{2} - 1) {}^{2} }} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \implies \: \sf{ \alpha {}^{2} + \beta {}^{2} = \sqrt{( {x}^{2} + 1) {}^{2} } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \therefore \: \underline{ \sf{ \: \alpha {}^{2} + \beta {}^{2} =x {}^{2} + 1 \: }}\: \: - - - - - - > (2)$

We have got by comparing that,

• $\: \: \: \: \underline{ \sf{ \: \alpha {}^{2} - \beta {}^{2} = 2x \: }} \: \: - - - - - - - - > (3)$

Adding Eq. (2) and Eq. (3), we get,

• $\: \: \bigstar \: \: \sf{2 \alpha {}^{2} = x {}^{2} + 1 + 2x}$

$\implies \: \sf{2 \alpha {}^{2} = (x + 1) {}^{2} } \\ \\ \implies \: \sf{ \alpha {}^{2} = \dfrac{(x + 1) {}^{2} }{2}} \\ \\ \implies \: \sf{ \alpha = \sqrt{ \dfrac{(x + 1) {}^{2} }{2} } } \\ \\ \: \: \: \: \: \boxed{\sf{ \therefore \: \: \alpha = \pm \: \dfrac{(x + 1 )}{ \sqrt{2} }}}$

Subtracting Eq. (3) from Eq. (2), we get,

• $\: \: \bigstar \: \: \sf{2 \beta {}^{2} = x {}^{2} + 1 - 2x}$

$\implies \: \sf{2 \beta {}^{2} = (x - 1) {}^{2} } \\ \\ \implies \: \sf{ \beta {}^{2} = \dfrac{(x - 1) {}^{2} }{2}} \\ \\ \implies \: \sf{ \beta = \sqrt{ \dfrac{(x - 1) {}^{2} }{2} } } \\ \\ \: \: \: \: \: \boxed{\sf{ \therefore \: \: \beta = \pm \: \dfrac{(x - 1 )}{ \sqrt{2} }}}$

Now,

Substituting the values of α and β in Eq. (1), we get,

• $\bigstar \: \: \sf{ \sqrt{2x + (x {}^{2} - 1)i} = \alpha + i \beta}$

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

Hence,

• The square root of the complex number 2x + (x² - 1)i is $\: \underline{\boxed{ \sf{ \: \pm \bigg \{ \frac{(x + 1)}{ \sqrt{2}} + i \: \frac{(x - 1)}{ \sqrt{2} } \bigg \} \: }}}$