Question #fe653

1 Answer
Jul 5, 2017

\sqrt{-3i}=\sqrt{3}i\sqrt{i}

Explanation:

Apply radical rule \sqrt{-a}=\sqrt{-1}\sqrt{a}, assuming a\ge \0

\sqrt{-3i}=\sqrt{-1}\sqrt{3i}

=\sqrt{-1}\sqrt{3i}

Apply imaginary rule: \sqrt{-1}=i

=i\sqrt{3i}

Apply radical rule root(n]{ab}=\root(n]{a}\root(n]{b} , assuming a\ge \0,\b\ge \0

\sqrt{3i}=\sqrt{3}\sqrt{i}

=\sqrt{3}i\sqrt{i}