Simplify $\displaystyle\frac{1-i}{2+3i}$, where $i^2 = -1.$
Multiplying the numerator and denominator by the conjugate of the denominator, we have \begin{align*}
\frac{1-i}{2+3i} \cdot \frac{2-3i}{2-3i} &= \frac{1(2) + 1(-3i) - i(2) - i(-3i)}{2(2) + 2(-3i) + 3i(2) -3i(3i)}\\
& = \frac{-1-5i}{13} \\
&= \boxed{-\frac{1}{13} - \frac{5}{13}i}.
\end{align*}
-\frac{1}{13}-\frac{5}{13}i
Level 5
Algebra
Problem: Simplify $\displaystyle\frac{1-i}{2+3i}$, where $i^2 = -1.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first multiply the numerator and denominator by the conjugate of the denominator, which is \(2 - 3i\). This step is crucial because it eliminates the imaginary part in the denominator. 

Calculating the numerator: 
\((1 - i)(2 - 3i) = 1 \cdot 2 + 1 \cdot (-3i) - i \cdot 2 - i \cdot (-3i) = 2 - 3i - 2i + 3i^2\). 
Since \(i^2 = -1\), this simplifies to \(2 - 3i - 2i - 3 = -1 - 5i\).

Next, calculate the denominator:
\((2 + 3i)(2 - 3i) = 2^2 - (3i)^2 = 4 - 9(-1) = 4 + 9 = 13\).

Now, combine the results:
\(\frac{-1 - 5i}{13}\).

This can be separated into real and imaginary parts:
\(-\frac{1}{13} - \frac{5}{13}i\).

Thus, the final answer is \(-\frac{1}{13} - \frac{5}{13}i\).<|END_OF_LATENT|> The final answer is $-\frac{1}{13}-\frac{5}{13}i$.