Question

3-i/1+i please answer in a +bi form

Answer 1

Now... let's multiply the denominator and numerator by the conjugate of the denominator as already suggested, which is just the same, but with a different sign in the middle, 1 - i.


$\bf \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b) \\\\\\ \textit{also recall that }\qquad i^2=(\sqrt{-1})(\sqrt{-1})\implies i^2=-1\\\\ -------------------------------$

$\bf \cfrac{3-i}{1+i}\cdot \cfrac{1-i}{1-i}\implies \cfrac{(3-i)(1-i)}{(1+i)(1-i)}\implies \cfrac{(3-i)(1-i)}{(1^2)-(i^2)} \\\\\\ \cfrac{3-3i-i+i^2}{1^2-(-1)}\implies \cfrac{3-3i-i+(-1)}{1+1}\implies \cfrac{3-1-3i-i}{2} \\\\\\ \cfrac{2-4i}{2}\implies \cfrac{\underline{2}(1-2i)}{\underline{2}}\implies 1-2i$