Answer
$\dfrac{3+5i}{1+i}=4+i$
Work Step by Step
$\dfrac{3+5i}{1+i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{3+5i}{1+i}=\dfrac{3+5i}{1+i}\cdot\dfrac{1-i}{1-i}=\dfrac{(3+5i)(1-i)}{1^{2}-i^{2}}=...$
$...=\dfrac{3-3i+5i-5i^{2}}{1-i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{3-3i+5i-5(-1)}{1-(-1)}=\dfrac{3+2i+5}{1+1}=\dfrac{8+2i}{2}=4+i$
