Existence of the square root in $\mathbb{C}$ I am stuck on the following Proposition:

Proposition: Show that for every $z \in \mathbb{C} \setminus (- \infty, 0]$ there exists exactly one $w \in \mathbb{C}$ such that $w^2=z$ and Re$(w)>0$

There are many things that puzzle me with this proposition:


*

*Most of it all that this proposition comes up in Real Analysis, yes real analysis. I haven't had complex Analysis yet but I am aware that many propositions in real analysis are not just to be taken over into complex analysis. Some of them however can be expanded such that they remain true in $\mathbb{C}$. 

*In $\mathbb{R}$ J.Appell shows in his book (Real Analysis 1) a more general form of the above Proposition using completeness of $\mathbb{R}$ and therefore nested intervals, it is a long yet very simple and constructive proof. From another book (Königsberger Analysis) I am aware that $\mathbb{C}$ is also complete, but I don't know how that helps me with the construction. 

*Furthermore I find it hard to intuitively grasp the set $Z= \mathbb{C} \setminus (-\infty,0]$, however I do know that $(-\infty,0] \subset \mathbb{R}$ which we might denote by $\mathbb{R_{-}}$ so we subtract from the complex numbers all negative real numbers including the zero. This is of course a set but I don't see how it's supposed to help me with the proposition.


Proof: Existence: Unfortunately I have absolutely no ideas here, is there a way to expand the real case into the set of $Z= \mathbb{C} \setminus (- \infty, 0]$ ?
Uniqueness: Let $h \in \mathbb{C}$ be another number such that $h^2=z$ with Re$(h)>0$ and lets assume that $w \neq h \implies w-h \neq 0$. So we obtain: $$w^2=h^2=z \implies w^2-h^2=\underbrace{(w-h)}_{\neq 0}(w+h)=0 \\ \implies w+h=0 $$
But $w$ and $h$ have both positive real values, so we get $w+h \neq 0$ therefore the contradiction and $w=h$
 A: 
The algebraic (quadratic polynomial, even) way: 

Some real numbers $a$ and $b$ are given, with conditions, such that $z=a+ib$ and one looks for solutions $(x,y)$ of $$w=x+iy,\qquad w^2=z,\qquad x\gt0,$$ that is, $$x^2-y^2=a,\qquad 2xy=b,\qquad x\gt0.$$ 
In particular, $x\gt0$ hence $$x=\sqrt{y^2+a},$$ and $y$ has the sign of $b$ with $$b^2=4x^2y^2=4(y^2+a)y^2=(2y^2+a)^2-a^2,$$ that is, $$y^2=\frac{-a+\sqrt{b^2+a^2}}2.$$
Finally, the unique solution $(x,y)$ is $$x=\sqrt{\frac{a+\sqrt{b^2+a^2}}2},\qquad y=\mathrm{sgn}(b)\,\sqrt{\frac{-a+\sqrt{b^2+a^2}}2}.$$ The only case where this could yield $x=0$ is if $a\lt0$ and $b=0$, fortunately this case was excluded from the start since it corresponds to $w$ on the halfline $\mathbb R_-$ in $\mathbb C$. Note also that $y$ involves the sign $\mathrm{sgn}(b)$, which is undefined if $b=0$ but this is no problem either since if $b=0$ then $a\gt0$ hence the square root yielding $y$ is $0$, thus one does not need to specify what would be  $\mathrm{sgn}(b)$ in this case.
A: OK. I'm going to assume the following theorem in real analysis: for any $r \ge 0$, there's a unique $s \ge 0$ with $s^2 = r$. In other words, square roots of positive numbers exist. 
Let $z = a + bi$; let $r = \sqrt{a^2+b^2}$ and $\theta = atan2(b, a)$. Then 
$$
a = r \cos \theta\\
b = r \sin \theta
$$
Let $s = \sqrt{r}$ and construct
$$
w = s \cos (\theta/2) + s \sin(\theta/2) i
$$
Then (as you can check by multiplication and an application of some double-angle formulas), we have $w^2 = z$. It's possible that $Re(w) < 0$, however; if so, replace $w$ with $-w$. Its square will also be $z$. 
So now I've covered existence. 
Uniqueness? Your proof looks good to me. 
By the way, I'm also assuming the definition of atan2, which is "atan2(s, c)" is the angle whose sine and cosine are positively proportional to $s$ and $c$, and for which the angle is between $-\pi$ and $\pi$; it's only defined for $s \ne0$ or $s = 0, c > 0$. My favorite definition is this:
$$
atan2(s, c) =
\begin{cases}
\arctan(s/c) & \text{if $c > 0$} \\
\pi + \arctan(s/c) & \text{if $c < 0, s> 0$} \\
-\pi + \arctan(s/c) & \text{if $c < 0, s< 0$}
\end{cases} 
$$
Since $\arctan$ can be defined by a convergent power series that you presumably already know about, this doesn't refer to anything you don't know. 
A: For all $z \in \mathbb{C} \ (- \infty , 0]$ write $z = \rho e^{i \theta}$ uniquely, where $\rho = |z| >0$ and $\theta \in (- \pi, \pi)$.
Then the square root of $z$ is $w =\sqrt{\rho} e^{i\frac{\theta}{2}}$. Note that $\frac{\theta}{2} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, hence $\Re w >0$.
A: The simplest and fastest way to see it the following:
Lemma. The function $s(z)=z^2$ is one-to-one and onto between
$$
U=\{z:\mathrm{Re}\,z>0\}\quad \mathit{and}\quad V=\mathbb C\smallsetminus(-\infty,0].
$$
Thus, for every $w\in\mathbb C\smallsetminus(-\infty,0]$, there is a unique $z$, with $\mathrm{Re}\,z>0$, such that $z^2=w$.
Proof of Lemma. 
a. $s$ is one-to-one. If $s(z_1)=s(z_2)$, then $z_1^2=z_2^2$ or $(z_1/z_2)^2=1$, which means that either $z_1=z_2$ or $z_1=-z_2$. But in the latter case only of the $z_1,z_2$ would have a positive real part. Contradiction.
b. $s$ is onto. Let $w\in\mathbb C\smallsetminus(-\infty,0]$. Then $w$ in polar coordinates
can be expressed as
$$
w=r\,\mathrm{e}^{i\vartheta},
$$
for some $r>0$, and $\vartheta\in(-\pi,\pi)$. In such case, $w=z^2$, where
$$
z=r^{1/2}\,\mathrm{e}^{i\vartheta/2},
$$
and as $\vartheta\in(-\pi/2,\pi/2)$, then $\mathrm{Re}\,z>0$.
