### Instruction: 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.### Response: I need to find the complex numbers that represent the vertices of the square.
Since the square is in counterclockwise order, I can assume that $0$ is the lower left vertex, $a$ is the lower right vertex, $b$ is the upper right vertex, and $c$ is the upper left vertex.
I can also assume that the side length of the square is $1$, since the problem does not specify otherwise.
Then, $a$ must be a real number, say $a = 1$.
$b$ must be a complex number with a real part of $1$ and an imaginary part of $1$, so $b = 1 + i$.
Similarly, $c$ must have a real part of $0$ and an imaginary part of $1$, so $c = i$.
Now, I have $0, 1, 1 + i, i$ as the vertices of the square.
To compute $\frac{ac + b^2}{ab}$, I need to use the rules of complex arithmetic.
$ac = 1 \cdot i = i$.
$b^2 = (1 + i)^2 = 1 + 2i - 1 = 2i$.
$ab = 1 \cdot (1 + i) = 1 + i$.
So, $\frac{ac + b^2}{ab} = \frac{i + 2i}{1 + i} = \frac{3i}{1 + i}$.
To simplify this fraction, I need to multiply the numerator and denominator by the complex conjugate of the denominator, which is $1 - i$.
This gives me $\frac{3i}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{3i - 3i^2}{1 - i^2} = \frac{3i + 3}{1 + 1} = \frac{3 + 3i}{2}$.
This is the answer in rectangular form.