### 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 want to find an expression for the fraction in terms of the real and imaginary parts of $a, b, c$.
I also notice that the fraction is invariant under rotations and translations, so I can assume without loss of generality that $0$ is the origin and $a$ is on the positive real axis.
Then, by the properties of a square, I know that $c = a(1 + i)$ and $b = a + ai$, where $i$ is the imaginary unit.
Substituting these values into the fraction, I get \[\frac{ac + b^2}{ab} = \frac{a^2(1 + i) + (a + ai)^2}{a(a + ai)}.\]
Simplifying the numerator, I get \[a^2(1 + i) + (a + ai)^2 = a^2(1 + i) + a^2 + 2a^2i - a^2 = 2a^2i.\]
Simplifying the denominator, I get \[a(a + ai) = a^2 + a^2i.\]
Dividing the numerator by the denominator, I get \[\frac{2a^2i}{a^2 + a^2i} = \frac{2i}{1 + i}.\]
To write this in rectangular form, I need to rationalize the denominator by multiplying the numerator and denominator by the complex conjugate of $1 + i$, which is $1 - i$.
Doing so, I get \[\frac{2i}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{2i - 2}{1 - i^2} = \frac{2(i - 1)}{2} = i - 1.\]
Therefore, the final answer is $i - 1$.