Question

divide:
2+4i/3+9i

Answer 1

\[\frac{2+4i}{3+9i}\] multiply by the conjugate of the denominator.
\[(a+bi)(a-bi)=a^2+b^2\] a real number so
\[(3+9i)(3-9i)=3^2+9^2=100\]

Answer 2

ooops
\[3^2+9^2=90\]

Answer 3

so you get
\[\frac{2+4i}{3+9i}=\frac{2+4i}{3+9i}\times \frac{3-9i}{3-9i}=\frac{(2+4i)(3-9i)}{90}\]

Answer 4

then
\[(2+4i)(3-9i)=42-6i\] so you get
\[\frac{42-6i}{90}\] or in standard form
\[\frac{7}{15}-\frac{1}{15}i\]