Question

Find the square root of the given complex number $\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}$
(a) $\pm \left\{ \dfrac{\left( a-b \right)+i\left( a+2b \right)}{2} \right\}$
(b) $\pm \left\{ \left( i+5b \right)+i\left( a-4b \right) \right\}$
(c) $-\sqrt{2}\left( a-b \right)$
(d) $\mp \left\{ \left( i+b \right)+i\left( a-b \right) \right\}$
(e) $\pm \left\{ \left( a+b \right)+i\left( a-b \right) \right\}$

Answer 1

- Hint:For solving this question, first we will derive the formula $4xy={{\left( x+y \right)}^{2}}-{{\left( x-y \right)}^{2}}$ by using the basic formulas like ${{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy$ and ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$ . After that, we will simplify the term $4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i$ and try to transform it into the whole square form. Then, we will apply the square root operator and select the correct option.

Complete step-by-step solution -

Given:
We have to solve the following term:
$\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}$
Now, before we proceed we should know the following formulas:
$\begin{align}
  & {{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy.............\left( 1 \right) \\
 & {{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy.............\left( 2 \right) \\
 & \left( {{x}^{2}}-{{y}^{2}} \right)=\left( x+y \right)\left( x-y \right).........\left( 3 \right) \\
 & i=\sqrt{-1}......................................\left( 4 \right) \\
\end{align}$
Now, subtract equation (1) from equation (2). Then,
$\begin{align}
  & {{\left( x+y \right)}^{2}}-{{\left( x-y \right)}^{2}}=\left( {{x}^{2}}+{{y}^{2}}+2xy \right)-\left( {{x}^{2}}+{{y}^{2}}-2xy \right) \\
 & \Rightarrow {{\left( x+y \right)}^{2}}-{{\left( x-y \right)}^{2}}=2xy+2xy \\
 & \Rightarrow {{\left( x+y \right)}^{2}}-{{\left( x-y \right)}^{2}}=4xy \\
 & \Rightarrow 4xy={{\left( x+y \right)}^{2}}-{{\left( x-y \right)}^{2}}....................................\left( 5 \right) \\
\end{align}$
Now, let $4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i=N$ and we have to find the value of $\sqrt{N}$ .
Now, we will use the above four formulas to simplify the term $N=4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i$ . So, we will use the formula from the equation (3) to write ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the expression of $N$ . Then,
$\begin{align}
  & N=4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i \\
 & \Rightarrow N=4ab-2\left( a+b \right)\left( a-b \right)i \\
\end{align}$
Now, we will use the formula from the equation (5) to write $4ab={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}$ in the above equation. Then,
$\begin{align}
  & N=4ab-2\left( a+b \right)\left( a-b \right)i \\
 & \Rightarrow N={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}-2\left( a+b \right)\left( a-b \right)i \\
\end{align}$
Now, from the equation (4), we know that value of iota $i=\sqrt{-1}$ so, ${{i}^{2}}=-1$ and we can use this, to write $-{{\left( a-b \right)}^{2}}={{i}^{2}}{{\left( a-b \right)}^{2}}$ in the above equation $N={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}-2\left( a+b \right)\left( a-b \right)i$ . Then,
$\begin{align}
  & N={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}-2\left( a+b \right)\left( a-b \right)i \\
 & \Rightarrow N={{\left( a+b \right)}^{2}}+{{i}^{2}}{{\left( a-b \right)}^{2}}-2\left( a+b \right)\left( a-b \right)i \\
 & \Rightarrow N={{\left( a+b \right)}^{2}}+{{\left( i\left( a-b \right) \right)}^{2}}-2\left( a+b \right)\left( i\left( a-b \right) \right) \\
\end{align}$
Now, we will use the formula from the equation (1) with $x=\left( a+b \right)$ and $y=i\left( a-b \right)$ to write $N={{\left[ \left( a+b \right)-i\left( a-b \right) \right]}^{2}}$ in the above equation. Then,
$\begin{align}
  & N={{\left( a+b \right)}^{2}}+{{\left( i\left( a-b \right) \right)}^{2}}-2\left( a+b \right)\left( i\left( a-b \right) \right) \\
 & \Rightarrow N={{\left[ \left( a+b \right)-i\left( a-b \right) \right]}^{2}} \\
\end{align}$
Now, we will apply the square root operator on both terms in the above equation. Then,
$\begin{align}
  & N={{\left[ \left( a+b \right)-i\left( a-b \right) \right]}^{2}} \\
 & \Rightarrow \sqrt{N}=\pm \left\{ \left( a+b \right)-i\left( a-b \right) \right\} \\
\end{align}$
Now, as per our assumption $N=4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i$ so, we can write $\sqrt{N}=\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}$ in the above equation. Then,
$\begin{align}
  & \sqrt{N}=\pm \left\{ \left( a+b \right)-i\left( a-b \right) \right\} \\
 & \Rightarrow \sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}=\pm \left\{ \left( a+b \right)-i\left( a-b \right) \right\} \\
\end{align}$
Now, from the above result, we conclude that the value of $\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}$ will be equal to the value of the term $\pm \left\{ \left( a+b \right)-i\left( a-b \right) \right\}$ .
Hence, the option (e) will be the correct option.

Note: Here, the student should first understand the given term $\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}$ and then, solve accordingly so, that our result will be matched by one of the given options. Moreover, we should remember the formula $4ab={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}$ and apply it directly and in such questions, we should try to stick to basic concepts and formulas like ${{i}^{2}}=-1$ , ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ . Then, we should solve it without any calculation mistakes and select the correct option.