# Write the complex number 8/(1+i) in standard form?

In the division of complex number you need first to change the denominator into a pure real number. To do that you multiply and divide by the complex conjugate of the denominator, in this case, $1 - i$. So:
$\frac{8}{1 + i} \cdot \frac{1 - i}{1 - i} = \frac{8 - 8 i}{1 + 1} = \frac{8}{2} - \frac{8}{2} i = 4 - 4 i$
Remember that: $\left(1 + i\right) \left(1 - i\right) = 1 - i + i - {i}^{2}$ but i^2=sqrt(-1))^2=-1
giving: $\left(1 + i\right) \left(1 - i\right) = 1 + 1 = 2$