How do you simplify #(8 i)/(2-i)#?

1 Answer
May 20, 2016

Answer:

#(8i)/(2-i)=-8/5+16/5i#

Explanation:

Given a complex number #a+bi#, the complex conjugate of that number is #a-bi#. A useful property is that the product of a complex number and its conjugate is a real number.

In this case, we will use that property, along with the fact that the conjugate of #2-i# is #2+i#, to remove the complex number from the denominator.

#(8i)/(2-i) = (8i(2+i))/((2-i)(2+i))#

#=(-8+16i)/5#
#=-8/5+16/5i#