Courses
Courses for Kids
Free study material
Free LIVE classes
More LIVE
Join Vedantu’s FREE Mastercalss

# Find the square root of $- 8 + 6i$ ?${\text{A}}{\text{.}} \pm {\text{(1 + 3i)}} \\ {\text{B}}{\text{.}} \pm {\text{(1 - 3i)}} \\ {\text{C}}{\text{.}} \pm {\text{(3 + i)}} \\ {\text{D}}{\text{.}} \pm {\text{(3 - i)}} \\$ Verified
360k+ views
Hint: In this type of question, where we have to find the square root of a complex number, the standard way is to assume that the square root of the given complex number is a new complex number which is x+iy and then square both sides. Solve the equation formed to get the value of the square root of the given complex number.

In the question, it is given a complex number -8+6i.
Because the number given is a complex number, so, we cannot directly find the value of the square root.
Let us first assume that the square root of a given complex number is x+iy.
$\therefore$ According to question, we can write:
$\sqrt { - 8 + 6i} = ({\text{x + iy)}}$ .
On squaring both side, we get:
$- 8 + 6i = {({\text{x + iy)}}^2}$ .
On solving the above equation, we get:
$- 8 + 6i = {{\text{x}}^2}{\text{ - }}{{\text{y}}^2} + 2i{\text{xy}}$ .
Now, equating the real and imaginary part on both side, we get:
$- 8 = {{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$ ------ (1)
And
$2{\text{xy = 6}} \\ \Rightarrow {\text{xy = }}\dfrac{6}{2} = 3 \\$ ----------------- (2)
We know that ${({\text{a + b)}}^2} = {\left( {{\text{a - b}}} \right)^2} + 4{\text{ab}}$ .

Therefore, we can write:
${({{\text{x}}^2}{\text{ + }}{{\text{y}}^2})^2} = {({{\text{x}}^2}{\text{ - }}{{\text{y}}^2})^2} + 4{\left( {{\text{xy}}} \right)^2}$ .
Putting the values from Equation 1 and 2, we get:
${({{\text{x}}^2}{\text{ + }}{{\text{y}}^2})^2} = {( - 8)^2} + 4{\left( 3 \right)^2} = 64 + 36 = 100 \\ \Rightarrow ({{\text{x}}^2} + {{\text{y}}^2}) = \pm \sqrt {100} = \pm 10 \\$
$\because$ x and y are real numbers. So, the sum of squares of x and y can never be negative.
So, the only solution is:
$({{\text{x}}^2} + {{\text{y}}^2}) = 10$ -----------(3)
On adding equation 1 and 3, we get:
$2{{\text{x}}^2} = 2 \\ \Rightarrow {{\text{x}}^2} = \dfrac{2}{2} = 1 \\ \Rightarrow {\text{x}} = \pm \sqrt 1 = \pm 1 \\$
Putting the value of x in equation 3, we get:
${1^2} + {{\text{y}}^2} = 10 \\ \Rightarrow {{\text{y}}^2} = 10 - 1 = 9 \\ \Rightarrow {\text{y = }} \pm \sqrt 9 = \pm 3. \\$
But, from equation 2:
${\text{xy = 3}}$ .
Since the product of x and y is positive. So, x and y can be either both positive or can both be negative.
Therefore, the square root of $- 8 + 6i = \pm (1 + 3i)$ i.e. (1+3i) and (-1-3i).

So, option A is correct.

Note: In this type of question the first step is to assume the square root of a given complex number as an unknown complex number and then square both sides to get an equation in x and y .After this use the algebraic identities to find the value of unknown parameter x and y. One point to be noted is that not all the value of x and y will give the required complex number. We have to take only those values which satisfy the remaining equation which in this case is xy=3.

Last updated date: 19th Sep 2023
Total views: 360k
Views today: 8.60k