Question

find square root -6-8i​

Answer 1

$\red{ \large \sf \underline{ \underline{ \: Given : \: \: \: }}}$

$\star \: \sf Complex \: number \: z = -6 - 8i \\ \\\star \: \sf Real \: part \: Re_{(z)} = - 6 \\ \\ \star \: \sf Imaginary \: part \: Im_{(z)} = - 8 < 0 \\ \\ \star \: \sf Modulus \: of \: z \: ( |z| ) = \sqrt{ {(a)}^{2} + {(b)}^{2} }$

$= \sqrt{ {( - 6)}^{2} + {( - 8)}^{2} } \\ \\ = \sqrt{36 + 64} \\ \\ = \sqrt{100} \\ \\ = 10$

Now ,

We know that , the square root of complex number (z) , if imaginary part is less than zero is given by :

$\large \sf\fbox{ \fbox{ \sqrt{z} = ± \sqrt{ \frac{ |z| + Re_{(z)}}{2} } - i \sqrt{ \frac{ |z| - Re_{(z)}}{2} } \: \: \: \: }}$

So ,

$\sf = ± \sqrt{ \frac{10 + ( - 6)}{2} } - i \sqrt{ \frac{10 - ( - 6)}{2} } \\ \\ \sf = ± \sqrt{ \frac{10 - 6}{2} } - i \sqrt{ \frac{10 + 6}{2} } \\ \\ = \sf ± \sqrt{ \frac{4}{2} } - i \sqrt{ \frac{16}{2} } \\ \\\sf = ± \sqrt{2} - i \sqrt{8}$

Hence , $\sf± \sqrt{2} - i \sqrt{8}$ is the square root of $\sf - 6 - 8i$