Question #fe653

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Answer 1 · 2017-07-05T01:45:55

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

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}#