Question

The square roots of$-7-24\sqrt{-1}$ are

Answer 1

Step-1: Assume a complex number to be the square root of the given complex number and form equations:

$-7-24\sqrt{-1}=-7-24i$, where $i=\sqrt{-1}$

Let $x+iy$ be the square of the complex number $-7-24i$

$\Rightarrow$ $-7-24i={\left(x+iy\right)}^2$

$=\left(x+iy\right)\left(x+iy\right)$

$=x^2+ixy+iyx+i^2{y}^2$

$\Rightarrow$ $-7-24i$ $=x^2-y^2+i\left(2xy\right)$ $\left[\because i^2=-1\right]$

Step-2: Solve the formed equations to find the value of $x$

On comparing the real and imaginary parts we get

$\Rightarrow$$x^2-y^2=-7$ $...\left(i\right)$ and $2xy=-24$$...\left(ii\right)$

$\Rightarrow$$x^2+y^2=\sqrt{{\left(x^2-y^2\right)}^2+4{\left(xy\right)}^2}$ $\left[\because a+b=\sqrt{{\left(a-b\right)}^2+4ab}\right]$

$=\sqrt{{\left(-7\right)}^2+{\left(-24\right)}^2}$

$=\sqrt{49+576}$

$\Rightarrow$$x^2+y^2=25$$...\left(iii\right)$

Adding $\left(i\right)$ and $\left(iii\right)$ we get

$2x^2=18$

$\Rightarrow$$x^2=9$

$\Rightarrow$ $x=\pm 3$

Step-3: Solve for the required value:

From equation $\left(ii\right)$, $2xy=-24\Rightarrow y=\frac{-12}{x}$

When $x=3$, $y=\frac{-12}{3}=-4$

When $x=-3$, $y=\frac{-12}{-3}=4$

The square root is $x+iy$.

Hence, substituting the appropriate values of $x$ and $y$ we get,

$x+iy=3-4i$ and $x+iy=-3+4i$

Hence, the square roots of the number $-7-24\sqrt{-1}$ are $-3+4i$ and $3-4i$.