We are given that $V = 1-i$ and $Z = 1+3i$, and we want to find $I$.
We can rearrange the equation $V = IZ$ to solve for $I$:
\[I = \frac{V}{Z}.\]
Substituting the given values, we have:
\[I = \frac{1-i}{1+3i}.\]
To simplify this complex fraction, we can multiply the numerator and denominator by the conjugate of the denominator:
\[I = \frac{(1-i)(1-3i)}{(1+3i)(1-3i)}.\]
Expanding and simplifying, we get:
\[I = \frac{1-4i+3i^2}{1-9i^2}.\]
Since $i^2 = -1$, we can simplify further:
\[I = \frac{1-4i-3}{1+9}.\]
Simplifying again, we have:
\[I = \frac{-2-4i}{10}.\]
Finally, we can divide both the numerator and denominator by 2 to get the simplified form:
\[I = \frac{-1-2i}{5}.\]
Therefore, $I = \boxed{-\frac{1}{5}-\frac{2}{5}i}$.The answer is: -\frac{1}{5}-\frac{2}{5}i
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Complex numbers are often used when dealing with alternating current (AC) circuits. In the equation $V = IZ$, $V$ is voltage, $I$ is current, and $Z$ is a value known as impedance. If $V = 1-i$ and $Z=1+3i$, find $I$.  Express your answer as a complex number in the form $a+bi$, where $a$ and $b$ are real numbers.

### Answer:
