Question

Find the square root of $-8+6i$ ?
$$\begin{gathered} \text{A}\text{.}\pm \text{(1 + 3i)} \\ \text{B}\text{.}\pm \text{(1 - 3i)} \\ \text{C}\text{.}\pm \text{(3 + i)} \\ \text{D}\text{.}\pm \text{(3 - i)} \end{gathered}$$

Answer 1

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.


Complete step-by-step answer:
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
$$\begin{gathered} 2\text{xy = 6} \\ \Rightarrow \text{xy = }{\displaystyle \frac{6}{2}}=3 \end{gathered}$$ ----------------- (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:
$$\begin{gathered} ({\text{x}}^2\text{ + }{\text{y}}^2{)}^2=(-8{)}^2+4{\left(3\right)}^2=64+36=100 \\ \Rightarrow ({\text{x}}^2+{\text{y}}^2)=\pm \sqrt{100}=\pm 10 \end{gathered}$$
$\because$ 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:
$$\begin{gathered} 2{\text{x}}^2=2 \\ \Rightarrow {\text{x}}^2={\displaystyle \frac{2}{2}}=1 \\ \Rightarrow \text{x}=\pm \sqrt{1}=\pm 1 \end{gathered}$$
Putting the value of x in equation 3, we get:
$$\begin{gathered} 1^2+{\text{y}}^2=10 \\ \Rightarrow {\text{y}}^2=10-1=9 \\ \Rightarrow \text{y = }\pm \sqrt{9}=\pm 3. \end{gathered}$$
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.