Let $0, a, b, c$ be the vertices of a square in counterclockwise order.  Compute
\[\frac{ac + b^2}{ab}.\]Enter your answer in rectangular form.
Here's a labelled picture of our square, with the vertices connected to the origin:
[asy]
import TrigMacros; 
size(180); 

pair O, A, B, C;

rr_cartesian_axes(-2, 8, -5, 7, complexplane = true, usegrid = false); 
O = (0,0); 
A = (3, -1); 
B = scale(sqrt(2))*rotate(45)*A; 
C = rotate(90)*A; 

draw(A--B--C--O--cycle); 
draw(O--B); 

dot("$a$", A, S); 
dot("$b$", B, E); 
dot("$c$", C, N); 
dot("$0$", O, SW); 
[/asy]

We know $b$ is a rotation of $a$ by $\pi/4$ around the origin, scaled by a factor of $\sqrt{2}$. That means that $b = \sqrt{2}e^{\pi i/4} a$, which becomes
\begin{align*} 
b &= \sqrt{2}(\cos (\pi/4) + i \sin(\pi/4))a \\
    &= \sqrt{2}\left( \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} i\right)a \\
    &= (1+i)a.
\end{align*}Therefore, $\frac{b}{a} = 1+i.$

Similarly, $c$ is a rotation of $b$ by $\pi/4$ around the origin, scaled by a factor of $\frac{1}{\sqrt{2}},$  That means $c = \frac{e^{\pi i/4}}{\sqrt{2}} b,$ which becomes
\[c = \frac{\sqrt{2}/2 + \sqrt{2}/2 \cdot i}{\sqrt{2}} b = \frac{1 + i}{2} b.\]Therefore, $\frac{c}{b} = \frac{1 + i}{2}.$

Then
\[\frac{ac + b^2}{ab} = \frac{c}{b} + \frac{b}{a} = \frac{1 + i}{2} + 1 + i = \boxed{\frac{3}{2} + \frac{3}{2} i}.\]