Question

Find the square root of $7-24i$.

Answer 1

Step 1. Solve for the square root.

The Imaginary number is a non-real number that is usually a multiple of $i$, which is the square root of $-1$.

The given number $7-24i$ is an imaginary number.

Let, $\sqrt{-7+24i}=x+iy$

On squaring, we get,

$-7+24i=\left(x+iy\right)\left(x+iy\right)$

$-7+24i=x^2+ixy+ixy+i^2{y}^2$

$-7+24i=\left(x^2-y^2\right)+2ixy$ $...\left[i^2=-1\right]$

Step 2: Equate the real and imaginary parts.

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

$\therefore$ $x^2+y^2=\sqrt{{\left(x^2-y^2\right)}^2+4x^2{y}^2}$

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

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

Adding , $\left(i\right)$ and $\left(ii\right)$

We get,

$\Rightarrow$ $2x^2=18$

$\Rightarrow$ $x^2=9$

$\Rightarrow$ $x=\pm 3$

And subtracting , $\left(ii\right)$ from $\left(i\right)$

We get,

$\Rightarrow$ $2y^2=32$

$\Rightarrow$ $y^2=16$

$\Rightarrow$ $y=\pm 4$

Since, $xy$ is positive, and have the same sign.

Therefore,

$\Rightarrow$ $x=3$

$\Rightarrow$ $y=4$

or

$\Rightarrow$ $x=-3$

$\Rightarrow$ $y=-4$

So,

$\Rightarrow$ $\sqrt{-7+24i}=\left(3+4i\right)or\left(-3-4i\right)$

Hence, the square root of $7-24i$ is $\left(3+4i\right)or\left(-3-4i\right)$.