What is the complex conjugate of #sqrt(-15)#?

1 Answer
George C.
Oct 29, 2015

#- sqrt(-15) = -i sqrt(15)#

Explanation:

If #x >= 0#, then #sqrt(x)# means the non-negative square root of #x#.

If #x < 0# then #sqrt(x) = i sqrt(-x)#

So #sqrt(-15) = i sqrt(15)#

The complex conjugate is formed by replacing #i# with #-i#, so the complex conjugate of #sqrt(-15) = i sqrt(15)# is #-sqrt(-15) = -i sqrt(15)#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
8783 views around the world
You can reuse this answer
Creative Commons License