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Question: Find the square root of \(8 - 6i\) . A.\( \pm \left( {3 - i} \right)\) B.\( \pm \left( {4 + 3i}...

Find the square root of 8−6i8 - 6i .
A.±(3−i) \pm \left( {3 - i} \right)
B.±(4+3i) \pm \left( {4 + 3i} \right)
C.±3(1−3i) \pm \sqrt 3 \left( {1 - 3i} \right)
D.None of these

Explanation

Solution

Hint : To carry out the square root of any number which is complex (a number having real and imaginary part) use the identity of x2−y2{x^2} - {y^2} and2xy2xy .

Complete step-by-step answer :
Here the given expression is: 8−6i8 - 6i
Where, the real part is 8 and the imaginary part is 6i.
The general form of the complex number is: x+iyx + iy
Taking the given expression equal to general form then we get,
⇒x+iy=8−6i\Rightarrow x + iy = \sqrt {8 - 6i}
Taking square to both sides in the given equation then we get,
⇒(x+iy)2=(8−6i)2 ⇒x2+(iy)2+2ixy=8−6i \Rightarrow {\left( {x + iy} \right)^2} = {\left( {\sqrt {8 - 6i} } \right)^2}\\\ \Rightarrow {x^2} + {\left( {iy} \right)^2} + 2ixy = 8 - 6i
Here value of i2{i^2} is -1 substituting the value in the equation then we get,
⇒x2+i2y2+2ixy=8−6i ⇒x2+(−1×y2)+2ixy=8−6i ⇒x2−y2+2ixy=8−6i \Rightarrow {x^2} + {i^2}{y^2} + 2ixy = 8 - 6i\\\ \Rightarrow {x^2} + \left( { - 1 \times {y^2}} \right) + 2ixy = 8 - 6i\\\ \Rightarrow {x^2} - {y^2} + 2ixy = 8 - 6i
Separating the real part and the imaginary part then we get,
⇒x2−y2=8,  2ixy=−6i ⇒x2−y2=8..............................(i) ⇒2xy=−6................................(ii) \Rightarrow {x^2} - {y^2} = 8,\;2ixy = - 6i\\\ \Rightarrow {x^2} - {y^2} = 8..............................(i)\\\ \Rightarrow 2xy = - 6................................(ii)
Taking equation (ii) then we get,
2xy=−6 xy=−3 x=−3y 2xy = - 6\\\ xy = - 3\\\ x = \dfrac{{ - 3}}{y}
Substituting the value of x in the equation (i) then we get,
⇒(−3y)2−y2=8 ⇒9−y4=8y2 ⇒y4+8y2−9=0 \Rightarrow {\left( {\dfrac{{ - 3}}{y}} \right)^2} - {y^2} = 8\\\ \Rightarrow 9 - {y^4} = 8{y^2}\\\ \Rightarrow {y^4} + 8{y^2} - 9 = 0
Solving the equation using the middle term splitting method then we get,
⇒y4+8y2−9=0 ⇒y4+9y2−y2−9=0 \Rightarrow {y^4} + 8{y^2} - 9 = 0\\\ \Rightarrow {y^4} + 9{y^2} - {y^2} - 9 = 0
Taking y2{y^2} common from first two terms and -1 common from last two terms then we get,
⇒y2(y2+9)−1(y2+9)=0 ⇒(y2+9)(y2−1)=0 \Rightarrow {y^2}\left( {{y^2} + 9} \right) - 1\left( {{y^2} + 9} \right) = 0\\\ \Rightarrow \left( {{y^2} + 9} \right)\left( {{y^2} - 1} \right) = 0
Getting the values by taking the square roots to left hand side then we get,
⇒(y2+9)=0,  (y2−1)=0 ⇒y2=−9,  y2=1 ⇒y=−9,  y=1 \Rightarrow \left( {{y^2} + 9} \right) = 0,\;\left( {{y^2} - 1} \right) = 0\\\ \Rightarrow {y^2} = - 9,\;{y^2} = 1\\\ \Rightarrow y = \sqrt { - 9} ,\;y = \sqrt 1
Since neglecting the negative value then we get,
y=1,−1y = 1, - 1
If we take y=1y = 1 then x=−3y=−31=−3x = \dfrac{{ - 3}}{y} = \dfrac{{ - 3}}{1} = - 3
If we take y=−1y = - 1 then x=−3−1=31=3x = \dfrac{{ - 3}}{{ - 1}} = \dfrac{3}{1} = 3
Hence the square root of the given complex expression 8−6i8 - 6i is ±(3−i) \pm \left( {3 - i} \right)
As there x is the real part and y is the imaginary party of the expression.
So, the correct answer is “Option A”.

Note : In this type of problem, we have to find the square of the complex number so, separate the real and imaginary part then carry out the square of the imaginary as well real part. This will make the square root of the given complex number.