Question

Find the square root of complex number 9 + 40i

Answer 1

Hint: To find the square root of a complex number, we will assume the root to be a + ib. Then we can compare it with the original number to find the values of a and b, which will give us the square root.

Complete step-by-step solution -
Let 9 + 40i = $(a+ib{)}^2$
We will simplify this equation, by proceeding as-
$9+40i=a^2+(ib{)}^2+2(ab)i$
We know that $i^2$= -1
$9+40i=a^2-b^2+2(ab)i$
By comparing the real and imaginary parts-
$a^2-b^2=9$............. (1)
$2ab=40$ ………………...(2)

Using equation (1) we can write that
$a^2=9+b^2$
$\text{a}=\pm \sqrt{9+{\text{b}}^2}$
Substituting this value in (2)-
$\pm 2\left(\sqrt{9+{\text{b}}^2}\right)\text{b}=40$
$\pm \text{b}\sqrt{9+{\text{b}}^2}=20$
Squaring both sides of the equation,
$b^2(9+b^2)=400$
$b^4+9b^2-400=0$
$b^4+25b^2-16b^2-400=0$
$(b^2-16)(b^2+25)$ = 0
$b^2$ = 16, $b^2$ = -25 (rejected because square cannot be negative)
b = 4, -4
Substituting the value of b in equation (1)-
$a^2$ - 16 = 9
$a^2$ = 25
a = 5, -5
So, the square roots of $9+40i$ are $5+4i$ and $-5-4i$.
This is the required answer.

Note: The most common mistake is that students only write one of the roots and do not write the other one. We should always take all the possible cases and reject them as and when required.