Question

The square root of $4ab - 2i\left( {{a^2} - {b^2}} \right)$:
1) $ \pm \left[ {\left( {a + b} \right) - i\left( {a - b} \right)} \right]$
2) $ \pm \left[ {\left( {a + b} \right) + i\left( {a - b} \right)} \right]$
3) $ \pm \left[ {\left( {ab} \right) + i\left( {a + b} \right)} \right]$
4) $ \pm \left[ {\left( {ab} \right) - i\left( {a + b} \right)} \right]$

Answer 1

Hint: We first let the root of the given number as $x + iy$. Then compare the real and imaginary part using the square of number $x + iy$ as $4ab - 2i\left( {{a^2} - {b^2}} \right)$. Find the value of ${x^2} + {y^2}$ using the formula ${\left( {{x^2} + {y^2}} \right)^2} = {\left( {{x^2} - {y^2}} \right)^2} + {\left( {2xy} \right)^2}$ and then solving the equations using elimination to find the value of $x$ and $y$. We get the final answer by substituting the values in $x + iy$.

Complete step by step answer:
Since the given number is a complex number, therefore the square root of this number will also be a complex number.
So, let the square root of the given number $4ab - 2i\left( {{a^2} - {b^2}} \right)$ as $x + iy$ . Thus we can write
${\left( {x + iy} \right)^2} = 4ab - 2i\left( {{a^2} - {b^2}} \right)$
On simplifying the above equation we get,
${x^2} - {y^2} - 2ixy = 4ab - 2i\left( {{a^2} - {b^2}} \right){\text{ }}\left( {\text{1}} \right)$
The part of the complex number without the $i$ as coefficient is called the real part of the complex number, and the part of the complex number with $i$ as its coefficient is called the imaginary part of the complex number.
On comparing the real and imaginary part of the equation $\left( 1 \right)$, we get
$
  {x^2} - {y^2} = 4ab \\
   - 2ixy = - 2i\left( {{a^2} - {b^2}} \right){\text{ }} \\
 $
On simplifying the equation we get
$
  {x^2} - {y^2} = 4ab{\text{ }}\left( {\text{2}} \right) \\
  xy = {a^2} - {b^2}{\text{ }}\left( {\text{3}} \right) \\
 $
Find the value of ${x^2} + {y^2}$using the formula ${\left( {{m^2} + {n^2}} \right)^2} = {\left( {{m^2} - {n^2}} \right)^2} + 4{m^2}{n^2}$
${\left( {{x^2} + {y^2}} \right)^2} = {\left( {{x^2} - {y^2}} \right)^2} + {\left( {2xy} \right)^2}$
Substituting the values from equation $\left( 2 \right){\text{ and }}\left( 3 \right)$ we get
$
  {\left( {{x^2} + {y^2}} \right)^2} = {\left( {4ab} \right)^2} + {\left( {2\left( {{a^2} - {b^2}} \right)} \right)^2} \\
   = 16{a^2}{b^2} + 4{\left( {{a^2} - {b^2}} \right)^2} \\
   = 16{a^2}{b^2} + 4\left( {{a^4} + {b^4} - 2{a^2}{b^2}} \right) \\
   = 8{a^2}{b^2} + 4{a^4} + 4{b^4} \\
   = {\left( {2{a^2} + 2{b^2}} \right)^2} \\
  {x^2} + {y^2} = 2{a^2} + 2{b^2}{\text{ }}\left( {\text{4}} \right) \\
 $
Adding equation $2$ and $4$we get
$
  {x^2} - {y^2} + {x^2} + {y^2} = 2{a^2} + 2{b^2} + 4ab{\text{ }} \\
  {\text{2}}{x^2} = 2{\left( {a + b} \right)^2} \\
  x = \pm \left( {a + b} \right) \\
 $
Substituting the $a + b$for $x$in the equation $1$.
$
  {\left( {a + b} \right)^2} - {y^2} = 4ab \\
   - {y^2} = 4ab - {a^2} - {b^2} - 2ab \\
   - {y^2} = - \left( {{a^2} + {b^2} - 2ab} \right) \\
  {y^2} = {\left( {a - b} \right)^2} \\
  y = \pm \left( {a - b} \right) \\
 $
Substituting the values of $x,y$ in the required complex number
$x + iy = \pm \left[ {a + b + i\left( {a - b} \right)} \right]$
Hence, option B is the correct answer.

Note: While calculating, students must remember the value of ${i^2}$ is $ - 1$. Substitute ${i^2} = - 1$ while solving ${\left( {x + iy} \right)^2}$. Calculation needs to be done carefully for these types of questions.