### Instruction: Simplify $\displaystyle\frac{1-i}{2+3i}$, where $i^2 = -1.$.### Response: So we want to simplify $\displaystyle\frac{1-i}{2+3i}$.
Right. Let's start by multiplying the numerator and denominator by the conjugate of $2+3i$, which is $2-3i$.
That gives us $\displaystyle\frac{(1-i)(2-3i)}{(2+3i)(2-3i)}$.
Now let's FOIL out the numerator and denominator.
So the denominator is $(2+3i)(2-3i) = 4 - 9i^2$.
And $i^2 = -1$, so the denominator is $4 - 9(-1) = 13$.
The numerator is $(1-i)(2-3i) = 2 - 3i - 2i + 3i^2$.
And $i^2 = -1$, so the numerator is $2 - 5i + 3(-1) = -1 - 5i$.
Our expression is now $\displaystyle\frac{-1-5i}{13}$.
And we can divide both the real and imaginary coefficients by 13 to get $\displaystyle\frac{-1}{13} - \displaystyle\frac{5}{13}i$.