Science:Math Exam Resources/Courses/MATH152/April 2017/Question A 03/Solution 1

Let $u/z=a+bi$. Then we have that
$(a+bi)z=u.$
We plug in z and u:
${\begin{aligned}(a+bi)(2-i)&=1+3i\\2a+2bi-ai-bi^{2}&=1+3i\\2a+b+(2b-a)i&=1+3i.\end{aligned}}$
We now equate the real and imaginary parts of both sides.
${\begin{aligned}2a+b&=1\\-a+2b&=3.\\\end{aligned}}$
We can multiply the second equation by 2:
${\begin{aligned}2a+b&=1\\-2a+4b&=6.\\\end{aligned}}$
Adding gives the equation
$5b=7.$ Therefore, we get that
$b=7/5.$
Plugging into either equation gives
$a=-1/5.$
Answer: $\color {blue}u/z=-{\frac {1}{5}}+{\frac {7}{5}}i$