standard form is a+bi where a and b are real numbers
remeber that i²=-1
ok
[tex]\frac{5}{3-15i}[/tex]
we got to get the i out of the denomenator
remember the differnce of 2 perfect squares where (a-b)(a+b)=a²-b²
so multiply the whole thing by [tex]\frac{3+15i}{3+15i}[/tex]
we get
[tex]\frac{5(3+15i)}{3^2-(15i)^2}=\frac{15+75i}{9-225(-1)}=\frac{15+75i}{9+225}=[/tex][tex]\frac{15+75i}{234}=\frac{15}{234}+\frac{75i}{234}=\frac{15}{234}+\frac{75}{234}i=\frac{5}{78}+\frac{25}{78}i[/tex]
in standard form, it is [tex]\frac{5}{78}+\frac{25}{78}i[/tex]
