How do you simplify #(2i) / (1 - i)#?

2 Answers
Mar 13, 2016

I found: #-1+i#

Explanation:

We can multiply and divide by the complex conjugate of the denominator, i.e., #1+i# to get:
#(2i)/(1-i)*(1+i)/(1+i)=(2i-2)/(1+1)=(2i-2)/2=#
remembering that: #i^2=-1#.
Finally:
#=2i/2-2/2=-1+i#

Mar 13, 2016

Multiply by the conjugate of the denominator to find that

#(2i)/(1-i) =i-1#

Explanation:

Given a complex number #a+bi# we call #a-bi# the complex conjugate of that number. One useful property is that the product of a complex number and its conjugate is a real number. We will use that to remove the complex number from the denominator by multiplying the numerator and denominator by the denominator's conjugate.

#(2i)/(1-i) = (2i(1+i))/((1-i)(1+i))#

#=(2(i-1))/(1+i-i+1)#

#=(2(i-1))/2#

#=i-1#