If $z_1=1-i$, $z_2=-2+4i$ and $a+ib=\dfrac {z_1.z_2}{{z_1}^{'}}$ then the values of $a$ and $b$ are : $$a. 2,-4$$ $$b. 4, \dfrac {5}{2}$$ $$c. 1, 0$$ $$d. 4, \dfrac {1}{2}$$
My Attempt ; $$a+ib=\dfrac {(1-i)(-2+4i)}{(1+I)}$$ $$=\dfrac {2+6i}{1+i}$$ $$=\dfrac {8+4i}{2}$$ $$=4+2i $$ On comparing, we have : $$a=4, b=2$$
If $z_1'$ means $\overline{z_1}$, then your answer is correct: the quotient is indeed $4+2i$.