-2
Explanation:[tex]17)\text{ }\frac{-4\text{ + 2i}}{2\text{ - i}}[/tex]
To simplify, we will use the conjugate of the denominator
conjugae of 2 - 1 = 2 + i
Multiply the conjugate of the denominator to the numerator and denominator:
[tex]\begin{gathered} \frac{-4\text{ + 2i}}{2\text{ - i}}\text{ }\times\frac{2+i}{2\text{ + i}} \\ =\text{ }\frac{(-4+2i)(2+i)}{(2-i)(2+i)} \\ \text{Expand:} \\ =\text{ }\frac{-4(2+i)+2i(2+i)}{(2-i)(2+i)} \end{gathered}[/tex][tex]\begin{gathered} =\text{ }\frac{-8-4i+4i+2i^2}{(2-i)(2+i)} \\ \\ \text{expand the denominator:} \\ =\text{ }\frac{-8-4i+4i+2i^2}{2(2+i)-i(2+i)} \\ =\text{ }\frac{-8-4i+4i+2i^2}{4+2i-2i-i^2} \end{gathered}[/tex][tex]\begin{gathered} In\text{ complex number, } \\ i^2\text{ = -1} \\ \text{substitute for i}^2 \\ =\text{ }\frac{-8-4i+4i+2i^2}{4+2i-2i-i^2}\text{ = }=\text{ }\frac{-8-4i+4i+2(-1)}{4+2i-2i-(-1)} \\ =\text{ }\frac{-8+0+2(-1)}{4+0-(-1)}\text{ = }\frac{-8-2}{4+1}\text{ } \\ =\text{ }\frac{-10}{5} \\ =\text{ -2} \end{gathered}[/tex]
The simplified answer is -2
