Answer
$\dfrac{3}{5}+\dfrac{1}{5}i$
Work Step by Step
Multiplying the numerator and the denominator by the conjugate of the denominator, the given expression, $
\dfrac{2}{3-i}
,$ is equivalent to
\begin{align*}\require {cancel}
&
\dfrac{2}{3-i}\cdot\dfrac{3+i}{3+i}
\\\\&=
\dfrac{2(3+i)}{3^2-i^2}
&\left( \text{use }(a+b)(a-b)=a^2-b^2 \right)
\\\\&=
\dfrac{2(3+i)}{9-i^2}
\\\\&=
\dfrac{2(3)+2(i)}{9-i^2}
&\left( \text{use Distributive Property } \right)
\\\\&=
\dfrac{6+2i}{9-i^2}
\\\\&=
\dfrac{6+2i}{9-(-1)}
&\left( \text{use }i^2=-1 \right)
\\\\&=
\dfrac{6+2i}{9+1}
\\\\&=
\dfrac{6+2i}{10}
\\\\&=
\dfrac{\cancel6^3+\cancel2^1i}{\cancel{10}^5}
&\left( \text{divide by }2 \right)
\\\\&=
\dfrac{3+i}{5}
\\\\&=
\dfrac{3}{5}+\dfrac{1}{5}i
.\end{align*}
Hence, in the form $a\pm bi,$ the given expression is equivalent to $
\dfrac{3}{5}+\dfrac{1}{5}i
$.
