If a+3i/2+ib=1-i, show that (5a-3b)=0.

if the given equation is :
$$\begin{gathered} a+\frac{3i}{2}+ib=1-i \\ a+\left(\frac{3}{2}+b\right)i=1-i \end{gathered}$$
comparing the real and imaginary part from the both sides of the equation , we have
$$\begin{gathered} a=1; \frac{3}{2}+b=-1 \\ b=-1-\frac{3}{2}=-\frac{5}{2} \end{gathered}$$
$$\begin{gathered} 2b=-5=-5*1 \\ 2b=-5a \\ \Rightarrow 5a+2b=0 \end{gathered}$$

please again go through the question and do get back to us.
since the result given is different from what we have calculated.

hope this helps you.