assuming ya mean [tex]\frac{3}{3-12i}[/tex]
and you want complex form as in a+bi
remember that [tex]i=\sqrt{-1}[/tex] so [tex]i^2=-1[/tex]
and remember conguates
we don't want imaginary numbers in the denomenator so
first sipmlify the farction
[tex]\frac{1}{1-4i}[/tex]
multiply it by [tex]\frac{1+4i}{1+4i}[/tex]
result is
[tex]\frac{1+4i}{(1)^2-(4i)^2}[/tex]
[tex]\frac{1+4i}{1-16i^2}[/tex]
[tex]\frac{1+4i}{1-16(-1)}[/tex]
[tex]\frac{1+4i}{1+16}[/tex]
[tex]\frac{1+4i}{17}[/tex]
[tex]\frac{1}{17}+\frac{4i}{17}[/tex]
in standard form (a+bi form) [tex]\frac{3}{3-12i}[/tex] is
[tex]\frac{1}{17}+\frac{4}{17}i[/tex]
