Solveeit Logo
Login

Question

Question: Find the square root of the given complex number \(\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}\)...

Find the square root of the given complex number 4ab−2(a2−b2)i\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) −2(a−b)-\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\\}

Explanation

Solution

- Hint:For solving this question, first we will derive the formula 4xy=(x+y)2−(x−y)24xy={{\left( x+y \right)}^{2}}-{{\left( x-y \right)}^{2}} by using the basic formulas like (x−y)2=x2+y2−2xy{{\left( x-y \right)}^{2}}={{x}^{2}}+{{y}^{2}}-2xy and (x+y)2=x2+y2+2xy{{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy . After that, we will simplify the term 4ab−2(a2−b2)i4ab-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:
4ab−2(a2−b2)i\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i}
Now, before we proceed we should know the following formulas:
(x−y)2=x2+y2−2xy.............(1) (x+y)2=x2+y2+2xy.............(2) (x2−y2)=(x+y)(x−y).........(3) i=−1......................................(4) \begin{aligned} & {{\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{aligned}
Now, subtract equation (1) from equation (2). Then,
(x+y)2−(x−y)2=(x2+y2+2xy)−(x2+y2−2xy) ⇒(x+y)2−(x−y)2=2xy+2xy ⇒(x+y)2−(x−y)2=4xy ⇒4xy=(x+y)2−(x−y)2....................................(5) \begin{aligned} & {{\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{aligned}
Now, let 4ab−2(a2−b2)i=N4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i=N and we have to find the value of N\sqrt{N} .
Now, we will use the above four formulas to simplify the term N=4ab−2(a2−b2)iN=4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i . So, we will use the formula from the equation (3) to write a2−b2=(a+b)(a−b){{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) in the expression of NN . Then,
N=4ab−2(a2−b2)i ⇒N=4ab−2(a+b)(a−b)i \begin{aligned} & N=4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i \\\ & \Rightarrow N=4ab-2\left( a+b \right)\left( a-b \right)i \\\ \end{aligned}
Now, we will use the formula from the equation (5) to write 4ab=(a+b)2−(a−b)24ab={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}} in the above equation. Then,
N=4ab−2(a+b)(a−b)i ⇒N=(a+b)2−(a−b)2−2(a+b)(a−b)i \begin{aligned} & 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{aligned}
Now, from the equation (4), we know that value of iota i=−1i=\sqrt{-1} so, i2=−1{{i}^{2}}=-1 and we can use this, to write −(a−b)2=i2(a−b)2-{{\left( a-b \right)}^{2}}={{i}^{2}}{{\left( a-b \right)}^{2}} in the above equation N=(a+b)2−(a−b)2−2(a+b)(a−b)iN={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}-2\left( a+b \right)\left( a-b \right)i . Then,
N=(a+b)2−(a−b)2−2(a+b)(a−b)i ⇒N=(a+b)2+i2(a−b)2−2(a+b)(a−b)i ⇒N=(a+b)2+(i(a−b))2−2(a+b)(i(a−b)) \begin{aligned} & 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{aligned}
Now, we will use the formula from the equation (1) with x=(a+b)x=\left( a+b \right) and y=i(a−b)y=i\left( a-b \right) to write N=[(a+b)−i(a−b)]2N={{\left[ \left( a+b \right)-i\left( a-b \right) \right]}^{2}} in the above equation. Then,
N=(a+b)2+(i(a−b))2−2(a+b)(i(a−b)) ⇒N=[(a+b)−i(a−b)]2 \begin{aligned} & 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{aligned}
Now, we will apply the square root operator on both terms in the above equation. Then,
\begin{aligned} & 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{aligned}
Now, as per our assumption N=4ab−2(a2−b2)iN=4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i so, we can write N=4ab−2(a2−b2)i\sqrt{N}=\sqrt{4ab-2\left( {{a}^{2}}-{{b}^{2}} \right)i} in the above equation. Then,
\begin{aligned} & \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{aligned}
Now, from the above result, we conclude that the value of 4ab−2(a2−b2)i\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 4ab−2(a2−b2)i\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=(a+b)2−(a−b)24ab={{\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 i2=−1{{i}^{2}}=-1 , a2−b2=(a+b)(a−b){{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.