# How do you simplify (18+3i)/(18-3i)?

Dec 13, 2015

$\setminus \frac{35 + 12 i}{37}$

#### Explanation:

First of all, you can simplify everything by a factor $3$:

$\setminus \frac{18 + 3 i}{18 - 3 i} = \setminus \frac{6 + i}{6 - i}$

Then, you can do something similar to razionalization with roots: multiply and divide by $6 + i$:

$\setminus \frac{6 + i}{6 - i} \setminus \frac{6 + i}{6 + i} = \setminus \frac{{\left(6 + i\right)}^{2}}{\left(6 + i\right) \left(6 - i\right)}$

Now use the fact that $\left(a + b\right) \left(a - b\right) = {a}^{2} - {b}^{2}$:

$\setminus \frac{{\left(6 + i\right)}^{2}}{{6}^{2} - {i}^{2}}$

Now remember that ${i}^{2} = - 1$:

 \frac{(6+i)^2}{6^2 +1} =\frac{36+12i-1}{37}=\frac{35+12i}{37}