So I assume
[tex]\frac{5}{4-3i}[/tex]
If so. Let us begin.
[tex]\mathrm{Multiply\:by\:the\:conjugate}\:\frac{4+3i}{4+3i} \ \textgreater \ \frac{5\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)} \ \textgreater \ \left(4-3i\right)\left(4+3i\right)[/tex]
[tex]\mathrm{Apply\:complex\:arithmetic\:rule}:\ \left(a+bi\right)\left(a-bi\right)=a^2+b^2[/tex]
[tex]Where\;a=4,\:b=-3 \ \textgreater \ 4^2+\left(-3\right)^2 \ \textgreater \ Refine \ \textgreater \ 25[/tex]
[tex]\frac{5\left(4+3i\right)}{25} \ \textgreater \ \mathrm{Cancel\:the\:common\:factor:}\:5 \ \textgreater \ \frac{4+3i}{5}[/tex]
[tex]Group\:the\:real\:part\:and\:the\:imaginary\:part\:of\:the\:complex\:number[/tex]
[tex]\frac{4}{5}+\frac{3}{5}i\;is\;our\;answer![/tex]
Hope this helps!