ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD=2AC. Its points D and M represent the complex numbers $1+i$ and  $2–i$respectively. And the complex number represented by  $A=a+\frac{3}{2}i,(a=\text{ real })$then
The value of ‘a’ = ----

Let A be (x, y)
It is given that BD=2AC$\Rightarrow$MD=2AM
Also DM is perpendicular to AM 
$\begin{array}{r}\Rightarrow (1-2{)}^2+(1+1{)}^2\\ =4\left[(\mathrm{x}-2{)}^2+(\mathrm{y}+1{)}^2\right]\end{array}$
and $\frac{\mathrm{y}+1}{\mathrm{x}-2}\cdot \frac{1+1}{1-2}=-1\Rightarrow 2(\mathrm{y}+1)=\mathrm{x}-2$
With $\mathrm{x}-2=2(\mathrm{y}+1),\left(1\right)\text{ gives }(\mathrm{y}+1{)}^2=1/4$
$\Rightarrow \mathrm{y}=-1/2,-3/2\Rightarrow \mathrm{x}=3,1$
$\Rightarrow$A represents z = 3 - i/2, or 1-3i/2
Alternative Sol. 
MD = 2AM and AM $\perp$DM i.e.  $\mathrm{\angle }\mathrm{AMD}=\mathrm{\pi }/2$
$\begin{array}{l}\Rightarrow \frac{\mathrm{z}-(2-\mathrm{i})}{(1+\mathrm{i})-(2-\mathrm{i})}=\frac{\mathrm{AM}}{\mathrm{MD}}\cdot {\mathrm{e}}^{\pm \frac{\mathrm{i\pi }}{2}}=\pm \mathrm{i}/2\\ \Rightarrow \mathrm{z}-(2-\mathrm{i})=\pm \frac{\mathrm{i}}{2}(-1+2\mathrm{i})\Rightarrow \mathrm{z}=3-\mathrm{i}/2\text{ or }1-3\mathrm{i}/2\end{array}$