Suppose $a$ and $b$ are real numbers, both not 0. Find real numbers $c$ and $d$ such that
$$ \frac{1}{a+bi} = c + di$$
I am not really sure what the question is asking me to do. Am I supposed to represent $c$ and $di$ both in terms of $a$ and $b$ since $a$ and $b$ are real numbers?
I multiplied the fraction by its conjugate, but that didn't give me any hints.
$$\frac{1}{a+bi} \bigg( \frac{a-bi}{a-bi}\bigg) = \frac{a-bi}{a^2+b^2}$$
A nudge in the correct direction would be helpful, thanks
Notice that $$ \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di. $$ Hence $$ c=\frac{a}{a^2+b^2}\,\, \text{ and }\,\,d=-\frac{b}{a^2+b^2}. $$