Simplify #sqrt(-64)#? Thanks

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Answer 1 · 2018-03-01T21:24:45

#8i#

#sqrt(-64) = sqrt(-1 * 64) #

# = sqrt(-1)*sqrt(64) #

# = sqrt(-1)*8 " " -> # since #i=sqrt(-1)#

# = 8i#

Answer 2 · 2018-03-01T21:26:16

#sqrt(-64)=8i#

On the set of complex numbers, the identity #sqrt(ab)=sqrtasqrtb# holds for any #a,binCC#.

So #sqrt(-64)=sqrt(-1)sqrt64=i8=8i#