Question

A complex number is given such that $x - iy = {( - 7 - 24i)^{\dfrac{1}{2}}}$ . What is the value of ${x^2} + {y^2}$ ?
A. 15
B. 25
C. -25
D. None of these

Answer 1

Hint: Iota, $i$ is a complex number such that $\sqrt i = - 1$ which gives us ${i^2} = - 1$ . Now, to solve the above question, use a relation which connects both ${({x^2} + {y^2})^2}$ and ${({x^2} - {y^2})^2}$ tomake the calculations further easier in evaluating the value of $({x^2} + {y^2})$ .

Complete step by step solution: 
The given number is: $x - iy = \sqrt {( - 7 - 24i)} $
On squaring both the sides, we get
${(x - iy)^2} = - 7 - 24i$
Opening the square,
${x^2} + {(iy)^2} - 2ixy = - 7 - 24i$
Now, the property of the complex number iota, $i$ is that ${i^2} = - 1$ .
Thus, proceeding further
${x^2} - {y^2} - 2ixy = - 7 - 24i$
On comparing both, the real part, and the imaginary part on both the sides we get:
${x^2} - {y^2} = - 7$ … (1)
And
$2xy = 24$ … (2)
Now, we will try to write $({x^2} + {y^2})$ in terms of $({x^2} - {y^2})$ .
We know that
${({x^2} + {y^2})^2} = {x^4} + {y^4} + 2{x^2}{y^2}$
Adding and subtracting $4{x^2}{y^2}$ on the right side of the above equation,
${({x^2} + {y^2})^2} = {x^4} + {y^4} + 2{x^2}{y^2} + 4{x^2}{y^2} - 4{x^2}{y^2}$
Simplifying further,
${({x^2} + {y^2})^2} = {x^4} + {y^4} - 2{x^2}{y^2} + 4{x^2}{y^2}$
As we know that $({x^2} - {y^2}) = {x^4} + {y^4} - 2{x^2}{y^2}$ . Hence, on substituting this we get:
${({x^2} + {y^2})^2} = {({x^2} - {y^2})^2} + 4{x^2}{y^2}$
Simplifying further,
${({x^2} + {y^2})^2} = {({x^2} - {y^2})^2} + {(2xy)^2}$
Substituting the values of $({x^2} - {y^2})$ from equation (1) and $(2xy)$ from equation (2) in the above equation,
${({x^2} + {y^2})^2} = {( - 7)^2} + {(24)^2} = 49 + 576$
Therefore, ${({x^2} + {y^2})^2} = 625$ .
Taking the square root, we get:
${x^2} + {y^2} = \pm 25$
As the sum of the squares of two numbers is always positive, therefore, ${x^2} + {y^2} = 25$ .
Thus, the correct option is B.

Note: On evaluating the square root of a variable, its value can be both positive and negative. For example, let’s say that ${x^2} = {a^2}$ . Then, on taking the square root one should always remember that $x = \pm a$ both. On analyzing the properties of $x$ , one can also further evaluate whether its value is positive and negative, just like we did in the question above.