Why is there only one point at infinity in the extended complex plane, but one in each direction in the real projective plane? It is a question that my friends recently discussed.
My opinion is that, by using one single point at infinity to form $\hat{\mathbb{C}}$, the behavior of functions such as $f(z)=z^2$ is just like that of a singularity at (near) the point $\infty$, and the phase of that point does not matter. However, in projective geometry we are concerned about the directions of lines, so we place one point at the 'end' of each line, and say that parallel lines meet there.
So the question is, is my view valid? Also, is there any other interesting reasons?
 A: There is a general construction that produces for any field $k$ and any $n\geq1$ the $n$-dimensional projective space $P(k,n)$. In the case $n=1$ this construction "adds a point $\infty$" to $k$. If $k={\mathbb R}$ we obtain the real projective line, where there is no distinction between $+\infty$ and $-\infty$. When $n=1$ and $k={\mathbb C}$ we obtain the extended complex plane  $P({\mathbb C},1)={\mathbb C}\cup\{\infty\}$ with just one point $\infty$. 
The definition of the real projective plane $P({\mathbb R},2)$ fully requires explaining the "general construction" referred to above: The projective plane $P({\mathbb R},2)$ is by definition the set of all one-dimensional subspaces of ${\mathbb R}^3$, i.e. the set of all lines to the origin of ${\mathbb R}^3$. This set is in bijective correspondence with the points of $S^2$ with antipodal points $x$ and $-x$ identified. When "doing geometry" in $P({\mathbb R},2)$ we rather work in ${\mathbb R}^2$ with a "line at infinity" added: To each (unoriented) direction in ${\mathbb R}^2$ corresponds a point on this line.
