#### Answer

$3-4i$

#### Work Step by Step

$\dfrac {-7+i}{-1-i}=\dfrac {7-i}{1+i}=\dfrac {\left( 7-i\right) \left( 1-i\right) }{\left( 1+i\right) \left( 1-i\right) }=\dfrac {7-7i-i+i^{2}}{1-i^{2}}=\dfrac {7-8i+i^{2}}{1-i^{2}}=\dfrac {7-8i+\left( -1\right) }{1-\left( -1\right) }=\dfrac {6-8i}{2}=3-4i$