Answer
$\dfrac{1-3i}{1+i}=-1-2i$
Work Step by Step
$\dfrac{1-3i}{1+i}$
Begin the evaluation of the quotient by multiplying the numerator and the denominator by the complex conjugate of the denominator:
$\dfrac{1-3i}{1+i}\cdot\dfrac{1-i}{1-i}=\dfrac{(1-3i)(1-i)}{1^{2}-i^{2}}=...$
Evaluate the operations indicated in the numerator and in the denominator:
$...=\dfrac{1-i-3i+3i^{2}}{1-i^{2}}=...$
Substitute $i^{2}$ by $−1$ and simplify:
$...=\dfrac{1-i-3i+3(-1)}{1-(-1)}=\dfrac{1-4i-3}{1+1}=\dfrac{-2-4i}{2}=...$
$...=-\dfrac{2}{2}-\dfrac{4}{2}i=-1-2i$
