The square root of $4 a b - 2 i (a^{2} - b^{2})$ :

\pm [(a + b) - i (a - b)]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm [(a + b) + i (a - b)]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\pm [(a b) + i (a + b)]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\pm [(a b) - i (a + b)]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\pm [(a + b) + i (a - b)]$
Let $(x + i y)^{2} = 4 a b - 2 i (a^{2} - b^{2})$
Comparing the real and imaginary terms,
$x^{2} - y^{2} = 4 a b$ ----------------------------------------(i)
$x y = a^{2} - b^{2}$ ---------------------------------------(ii)

Comparing the magnitude,
$x^{2} + y^{2} = \sqrt{4 a^{4} + 4 b^{4} + 8 a^{2} b^{2}}$
$\Rightarrow x^{2} + y^{2} = 2 a^{2} + 2 b^{2}$ -(iii)

Solving (i) and (iii),
$x^{2} = a^{2} + b^{2} + 2 a b$ and $y^{2} = a^{2} + b^{2} - 2 a b$

Hence,
$x = a + b$ , $y = a - b$
or $x = - (a + b)$ , $y = b - a$
Hence, option B is correct

Suggest Corrections

Similar questions

{(\frac{a}{b})}^{2} + {(\frac{b}{a})}^{2} - \frac{4}{i} (\frac{a}{b} - \frac{b}{a}) - 6

If $(a^{2} + b^{2}) = 1 \frac{1 + b + i a}{(1 + b - i a)}$

a^{2} + b^{2} = 1

\frac{1 + b + i a}{1 + b - i a} =

| \to A \times \to B | = \sqrt{3} \to A . \to B,

| \to A + \to B |

I_{A}

I_{B}

A & B

A & B

I_{A}

I_{B}

Join BYJU'S Learning Program