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

The square root of 4ab−2i(a2−b2)4ab - 2i\left( {{a^2} - {b^2}} \right):

  1. ±[(a+b)−i(a−b)] \pm \left[ {\left( {a + b} \right) - i\left( {a - b} \right)} \right]
  2. ±[(a+b)+i(a−b)] \pm \left[ {\left( {a + b} \right) + i\left( {a - b} \right)} \right]
  3. ±[(ab)+i(a+b)] \pm \left[ {\left( {ab} \right) + i\left( {a + b} \right)} \right]
  4. ±[(ab)−i(a+b)] \pm \left[ {\left( {ab} \right) - i\left( {a + b} \right)} \right]
Explanation

Solution

Hint: We first let the root of the given number as x+iyx + iy. Then compare the real and imaginary part using the square of number x+iyx + iy as 4ab−2i(a2−b2)4ab - 2i\left( {{a^2} - {b^2}} \right). Find the value of x2+y2{x^2} + {y^2} using the formula (x2+y2)2=(x2−y2)2+(2xy)2{\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 xx and yy. We get the final answer by substituting the values in x+iyx + 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(a2−b2)4ab - 2i\left( {{a^2} - {b^2}} \right) as x+iyx + iy . Thus we can write
(x+iy)2=4ab−2i(a2−b2){\left( {x + iy} \right)^2} = 4ab - 2i\left( {{a^2} - {b^2}} \right)
On simplifying the above equation we get,
x2−y2−2ixy=4ab−2i(a2−b2) (1){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 ii as coefficient is called the real part of the complex number, and the part of the complex number with ii as its coefficient is called the imaginary part of the complex number.
On comparing the real and imaginary part of the equation (1)\left( 1 \right), we get
x2−y2=4ab \-2ixy=−2i(a2−b2)   {x^2} - {y^2} = 4ab \\\ \- 2ixy = - 2i\left( {{a^2} - {b^2}} \right){\text{ }} \\\
On simplifying the equation we get
x2−y2=4ab (2) xy=a2−b2 (3)  {x^2} - {y^2} = 4ab{\text{ }}\left( {\text{2}} \right) \\\ xy = {a^2} - {b^2}{\text{ }}\left( {\text{3}} \right) \\\
Find the value of x2+y2{x^2} + {y^2}using the formula (m2+n2)2=(m2−n2)2+4m2n2{\left( {{m^2} + {n^2}} \right)^2} = {\left( {{m^2} - {n^2}} \right)^2} + 4{m^2}{n^2}
(x2+y2)2=(x2−y2)2+(2xy)2{\left( {{x^2} + {y^2}} \right)^2} = {\left( {{x^2} - {y^2}} \right)^2} + {\left( {2xy} \right)^2}
Substituting the values from equation (2) and (3)\left( 2 \right){\text{ and }}\left( 3 \right) we get
(x2+y2)2=(4ab)2+(2(a2−b2))2 =16a2b2+4(a2−b2)2 =16a2b2+4(a4+b4−2a2b2) =8a2b2+4a4+4b4 =(2a2+2b2)2 x2+y2=2a2+2b2 (4)  {\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 22 and 44we get
x2−y2+x2+y2=2a2+2b2+4ab  2x2=2(a+b)2 x=±(a+b)  {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+ba + bfor xxin the equation 11.
(a+b)2−y2=4ab \-y2=4ab−a2−b2−2ab \-y2=−(a2+b2−2ab) y2=(a−b)2 y=±(a−b)  {\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,yx,y in the required complex number
x+iy=±[a+b+i(a−b)]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 i2{i^2} is −1 - 1. Substitute i2=−1{i^2} = - 1 while solving (x+iy)2{\left( {x + iy} \right)^2}. Calculation needs to be done carefully for these types of questions.