Simplification of an unsolveable denominator. I am stuck on the problem:
Simplify : $\displaystyle \frac{3}{(2 + i)}$
Any explanations?
Thanks
 A: 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.
A: multiply both numerator and denominator by $(2-i)$. You will get $\frac{3(2-i)}{(2+i)(2-i)}=\frac{6-3i}{4-i^2}=\frac{6-3i}{4+1}=\frac{6-3i}{5}$
A: 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.
A: 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. 
A: Multiply by ${2-i \over 2-i}$
to get ${6-3i\over 5}$.
