# Simplification of an unsolveable denominator.

I am stuck on the problem:

Simplify : $\displaystyle \frac{3}{(2 + i)}$

Any explanations?

Thanks

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They probably want you to get it in a form without $i$ in the denominator, which you do by multiplying top and bottom by the complex conjugate. For a complex number $c = a+bi$ (here $2+i$), the complex conjugate is $\bar c = a-bi$. This has the property that $c \bar c = a^2 + b^2$ is a real number.

So multiplying top and bottom of your fraction by the complex conjugate of $2+i$ will give you a fraction which has a real number in the denominator, which is often considered preferable.

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Thanks Wikipedia! Besides my previous lack of knowledge of the complex conjugate, you gave a very informative answer. Thanks! – Cole Weis Jan 23 '13 at 18:31

multiply both numerator and denominator by $(2-i)$. You will get $\frac{3(2-i)}{(2+i)(2-i)}=\frac{6-3i}{4-i^2}=\frac{6-3i}{4+1}=\frac{6-3i}{5}$

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Thanks for being straight-forward. – Cole Weis Jan 23 '13 at 18:28

In general, if you're dividing by a complex number $a+bi$, where $a,b$ are real and $b\neq 0$, then you need to multiply numerator and denominator by the complex conjugate $a-bi$. This will give you a real number on the bottom. At that point, simplify as far as you can.

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I am starting to see that... Thanks! – Cole Weis Jan 23 '13 at 18:38

You can multiply and divide by $2-i$ :) in general, to write a complex number in a $a+ib$ form you can multiply and divide by the conjugate of the denominator.

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Qn: how can i delete my answer if other n+1 people wrote the same thing in the meanwhile?? – pppqqq Jan 23 '13 at 18:24
Sorry, my question was too unexotic. – Cole Weis Jan 23 '13 at 18:37

Multiply by ${2-i \over 2-i}$ to get ${6-3i\over 5}$.

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Yepperdoodle sir. – Cole Weis Jan 23 '13 at 20:29