Closure of the set in complex analysis... Here's the problem:

In each case, sketch the closure of the set:
(a) $−π < arg(z) < π$, $(z \neq 0)$
(b) $|Re (z)| < |z|$
(c) $Re(\frac{1}{z}) \leq \frac{1}{2}$
(d) $Re(z^2) > 0$

I've been having trouble with understanding the elementary Topology introduced in our book regarding the Complex plane. So here's what I've come to the conclusion of for (a):

"Solution" to (a)
We know that in polar form, $z=x+iy$ becomes $z=re^{i\theta}$, where
$$r=|z|=\sqrt{x^2+y^2}$$
$$arg(z)=Arg(z)+2\pi k$$
where $k \in \mathbb{N}$ and $Arg(z)=\theta$. It is given that $arg(z) \in (-\pi,\pi)$, so we must have $\theta \in (-\pi,\pi)$. Now, plugging in our endpoints of $\theta$, we end up with
$$re^{-i\pi}=r(cos(-\pi)+isin(-\pi))=r(1)=r$$
$$re^{i\pi}=r(cos(\pi)+isin(\pi))=r(-1)=-r.$$
We know that $z \neq 0$ $\implies r \neq 0$, so it would appear that we're closed in the entire complex plane except for at $0$.

I have no idea if this is correct, so if anyone could lead me down the right track, I'd be much obliged!
 A: So in the below problems I am using a varying degree of rigor.  Of course all of these arguments could be made more rigorous, but it is an exercise in tedium - the essential idea is that if you have a region bounded in $\mathbb{R}^2$, its boundaries, i.e. the lines that enclose it, must be part of its closure.  Once that is understood, the reast is just basic manipulation of complex numbers.  
Suppose $z = x + iy$.
(b) $|Re(z)| < |z|$ iff $|x| <\sqrt{x^2 + y^2}$ iff $ x^2 < x^2 + y^2$ iff $ 0 < y^2$ iff $0 \neq y$.  Hence this is the entire plane less the real axis. The closure, then, is the entire plane, since for any point in $\mathbb{C}$ of the form $(x, 0)$, we can pick elements of our set arbitrarily close (to make this rigorous consider points of the form $(x,y)$.  Then $|(x,y) - (x,0) | =\sqrt{y^2} = |y|$ which can be made less than any epsilon.)
(c)  Note that $Re(\frac{1}{z}) \leq \frac{1}{2}$ if and only if $Re( \frac{x - iy}{x^2 + y^2}) \leq \frac{1}{2}$ and thus if and only if $\frac{x}{x^2+y^2} \leq \frac{1}{2}$. This is only true if
$$
0 \leq y^2 + x^2 - 2x \iff 1 \leq (x-1)^2 + y^2.
$$ This is the entire plane, less the interior of the circle given by $1 = (x-1)^2 + y^2$.  Since this interior is open (this is easy to prove - just pick any point in the circle and select neighborhood that is a subset of the circle), it follows that its complement must be closed, and thus that our set it closed automatically and we must only draw the set itself.
(d) $Re(z^2) > 0$ if and only if $x^2 - y^2 > 0$.  This is the case if and only if $|x| > |y|$.  I'll let you graph this yourself, but it is clear that the boundary of this set is given by $y = \pm |x|$. 
