How do you divide #6/(-4i)#?

1 Answer
Aug 3, 2016

Answer:

#6/(-4i)=3/2i#

Explanation:

The complex conjugate of a complex number #a+bi#, denoted #bar(a+bi)#, is given by #bar(a+bi) = a-bi#. A useful property is that the product of any number and its conjugate is a real number, that is,

#a+bi*(bar(a+bi)) in RR#

We will use that property to remove the complex part from the denominator by multiplying the numerator and denominator by the conjugate of the denominator.

#6/(-4i) = (6*(bar(-4i)))/(-4i*(bar(-4i)))#

#=(6*(4i))/(-4i*(4i))#

#=(24i)/(16)#

#=3/2i#