How do you simplify #8/(1+i)# and write the complex number in standard form?

1 Answer
Jul 11, 2016

Answer:

#4 -4i#

Explanation:

Recall that we can multiply the top and bottom of a fraction by the same thing without changing the value of the fraction. In this case we will multiply by the complex conjugate of the complex number on the denominator.

For #z = 1+i# we have #bar(z) = 1 - i#

#8/(1+i) * (1-i)/(1-i) = (8-8i)/(1-i^2)#

Remember that #i^2 = -1# so we have

#(8-8i)/2 = 4 - 4i#