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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

Evan Kim
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1 Answers1

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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 $$ c=\frac{a}{a^2+b^2}\,\, \text{ and }\,\,d=-\frac{b}{a^2+b^2}. $$

sam wolfe
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    Oh, and I think the reason you can do this is because both sides of the equation are now in the format $(a + bi)$ so thats why you can get values of $c$ and $d$? – Evan Kim Aug 25 '19 at 23:53
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    Yes, that is correct. You compare the respective real and imaginary parts of both sides. – sam wolfe Aug 25 '19 at 23:54