Why is $\frac 25$ the real part of $\frac{1}{2+i}$? According to Wolfram Alpha, Re(1/(2+i))=2/5.  
How did it calculate that?
 A: $Re(z) = \frac 12 (z + \bar z)$
$\frac 1 {2+i} + \frac 1 {2-i} = \frac{2-i+2+i}{(2+i)(2-i)} = \frac 4 5$
A: $$\frac{1(2-i)}{(2+i)(2-i)} = \frac{2-i}{5} = \underbrace{\frac{2}{5}}_\text{real part} + i\cdot \overbrace{(-\frac{1}{5})}^\text{imaginary part}$$
A: Multiply numerator and denominator by $2-i$:
$$\frac 1{2+i}\cdot \frac{2-i}{2-i} = \frac{2-i}{4-(-1)} = \frac{2-i}{5} = \frac 25 - \frac 15\cdot i$$
Edit for more explanation as to why this strategy works. 
We multiply numerator and denominator by the conjugate of $2+ i$, which is $2+i$, to remove the imaginary number from the denominator. Doing so gives us a real-valued  denominator: $$(2 + i)(2-i) = 2^2 - (i^2) = 4 - (-1) = 5.$$ Having multiplied by $\frac{2-i}{2-i}=1$, that transforms the numerator to $1\cdot (2-i) = 2- i$. Then all that remained was to transform the fraction $\dfrac{2-i}{5}$ into its real and imaginary components.
The same strategy works for any fraction of the form $\dfrac{z_1}{z_2},\,$ where $z_1, z_2 \in \mathbb C$ and $z_2 = a + bi$ where $a, b \in \mathbb R, b\neq 0$. We simply multiply numerator and denominator by $\overline{z_2}$. The denominator is then $z_2\cdot \overline{z_2} = |z| = a^2 + b^2$
A: The unwritten goal for all of these answers is to write the complex number in the form
$$(\textrm{real number}) + (\textrm{another real number})i$$
The first real number is the real part and the second real number is the imaginary part.
So if you have a complex number $a+bi$ (with $a$ and $b$ real) as the denominator of a fraction, you can multiply top and bottom of the fraction by the so-called conjugate $a-bi$ to get rid of that. This works because it is always true that $(a+bi)(a-bi) = a^2+b^2$ is real.
Then you end up with a complex number in the numerator, which is easier to deal with. For example, suppose we wanted to find the real and imaginary parts of $\frac{1+3i}{2+i}$. Then we would write
$$\frac{1+3i}{2+i} = \frac{1+3i}{2+i}\cdot\left(\frac{2-i}{2-i}\right)$$
$$=\frac{(1+3i)(2-i)}{(2+i)(2-i)}$$
$$=\frac{2-i +6i+3}{4+1}$$
$$= \frac{5+5i}{5}$$
$$ = \left(\frac55\right) + \left(\frac55\right)i$$
$$= 1+i$$
So in this case, the real part is $1$ and the imaginary part also happens to be $1$.
A: $$\frac{1}{2+i}=\frac{2-i}{(2+i)(2-i)}=\frac{2-i}{5}=\frac{2}{5}-\frac{1}{5}i$$
A: $$
\frac{1}{2+i} = \frac{2-i}{(2+i)(2-i)}
$$
A: since $\frac{1}{2+i}=\frac{2-i}{4+1}=\frac{2-i}{5}$
A: $\frac{1}{a+bi}=\frac{1}{a+bi}\frac{a-bi}{a-bi}=\frac{a}{a^2+b^2}+i\cdot bla$
$a=2,b=1\implies \frac{a}{a^2+b^2}=\frac{2}{2^2+1^2}=\frac{2}{5}$
