Answer
$\dfrac{6}{5}+\dfrac{8i}{5}$
Work Step by Step
Multiply the numerator and the denominator by the conjugate of $3-4i$.
$\dfrac{10}{3-4i}=\dfrac{10}{3-4i}\cdot \dfrac{3+4i}{3+4i}$
Use special formula $(a+b)(a-b)=a^2-b^2$ in the denominator.
$=\dfrac{10(3+4i)}{(3)^2-(4i)^2}$
$=\dfrac{10(3+4i)}{9-16i^2}$
Use $i^2=-1$.
$=\dfrac{10(3+4i)}{9-16(-1)}$
Simplify.
$\require{cancel}
=\dfrac{10(3+4i)}{9+16}\\
=\dfrac{10(3+4i)}{25}\\
=\dfrac{\cancel{10}^2(3+4i)}{\cancel{25}^5}\\
=\dfrac{2(3+4i)}{5}$
Use distributive property in the numerator.
$=\dfrac{6+8i}{5}$
$=\dfrac{6}{5}+\dfrac{8i}{5}$
Hence, the solution in the standard form is $\dfrac{6}{5}+\dfrac{8i}{5}$.
