Question: User: z=1+i−1+3i−1−2i3+i
1. **State the problem:** Simplify the expression $$z=\frac{1+i}{-1+3i} - \frac{-1-2i}{3+i}$$.
2. **Recall the formula:** To simplify complex fractions, multiply numerator and denominator by the conjugate of the denominator to remove imaginary parts from the denominator.
3. **Simplify the first fraction:**
$$\frac{1+i}{-1+3i} \times \frac{-1-3i}{-1-3i} = \frac{(1+i)(-1-3i)}{(-1)^2 - (3i)^2} = \frac{(1)(-1) + (1)(-3i) + (i)(-1) + (i)(-3i)}{1 - (-9)}$$
$$= \frac{-1 - 3i - i - 3i^2}{1 + 9} = \frac{-1 - 4i - 3(-1)}{10} = \frac{-1 - 4i + 3}{10} = \frac{2 - 4i}{10}$$
4. **Simplify the fraction:**
$$\frac{2 - 4i}{10} = \frac{\cancel{2}(1 - 2i)}{\cancel{2}5} = \frac{1 - 2i}{5}$$
5. **Simplify the second fraction:**
$$\frac{-1-2i}{3+i} \times \frac{3 - i}{3 - i} = \frac{(-1-2i)(3 - i)}{3^2 - i^2} = \frac{-3 + i - 6i + 2i^2}{9 - (-1)}$$
$$= \frac{-3 - 5i + 2(-1)}{10} = \frac{-3 - 5i - 2}{10} = \frac{-5 - 5i}{10}$$
6. **Simplify the fraction:**
$$\frac{-5 - 5i}{10} = \frac{\cancel{-5}(1 + i)}{\cancel{-5}(-2)} = \frac{1 + i}{-2} = -\frac{1 + i}{2}$$
7. **Combine the two results:**
$$z = \frac{1 - 2i}{5} - \left(-\frac{1 + i}{2}\right) = \frac{1 - 2i}{5} + \frac{1 + i}{2}$$
8. **Find common denominator and add:**
$$z = \frac{2(1 - 2i)}{10} + \frac{5(1 + i)}{10} = \frac{2 - 4i + 5 + 5i}{10} = \frac{7 + i}{10}$$
**Final answer:**
$$z = \frac{7 + i}{10}$$