Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 02/Solution 1

Following the hint, we expand the product using the distributive property:

$zu=(1+3i)(1-i)=1-i+3i-3i^{2}=1+2i-3i^{2}.$

Then, using the fact that $i^{2}=-1,$ we get

$zu=1+2i-3i^{2}=1+2i+3=\color {blue}{4+2i}.$