Answer
$\dfrac{8}{1+\dfrac{2}{i}}=\dfrac{8}{5}+\dfrac{16}{5}i$
Work Step by Step
$\dfrac{8}{1+\dfrac{2}{i}}$
Evaluate the sum on the denominator:
$\dfrac{8}{1+\dfrac{2}{i}}=\dfrac{8}{\dfrac{2+i}{i}}=...$
Evaluate the division:
$...=\dfrac{8i}{2+i}=...$
Multiply the numerator and in the denominator of the fraction by the complex conjugate of the denominator:
$...=\dfrac{8i}{2+i}\cdot\dfrac{2-i}{2-i}=\dfrac{8i(2-i)}{2^{2}-i^{2}}=\dfrac{16i-8i^{2}}{4-i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{16i-8(-1)}{4-(-1)}=\dfrac{8+16i}{4+1}=\dfrac{8+16i}{5}=\dfrac{8}{5}+\dfrac{16}{5}i$
