To construct that branch of log $z$, you just define the half-parabola to be the branch cut. This would mean log $z$ = Log|$z$| + $i\theta$ where $\theta$ equals the value of arg $z$ between $\frac{\pi}{2}$ and $2\pi$ for $z$ in the second, third, or fourth quadrant, the value of arg $z$ between $0$ and $\frac{\pi}{2}$ for $z$ in the first quadrant above the half parabola, and the value of arg $z$ between $2\pi$ and $\frac{5\pi}{2}$ for z in the first quadrant below the half parabola.
Explicitly, you can solve for $\theta$ in terms of $r$ where $z=re^{i\theta}$. By the equation of the half parabola, $\frac{y}{x}=\frac{1}{y}$. Then, $\theta=\arctan(\frac{y}{x})=\arctan(\frac{1}{y})$.
$r=\sqrt{x^2+y^2}=\sqrt{x^2+x}$ , then $x^2+x-r^2=0$. By the quadratic formula,
$$x=\frac{-1\pm\sqrt{1+4r^2}}{2}$$
You choose the positive square root because you want the upper half parabola so,
$$y=\sqrt{\frac{\sqrt{1+4r^2}-1}{2}}$$
Therefore, the entire branch would be defined as
$$\log z = \mbox{Log }|z| + i\theta , \mbox{where } \theta = \arctan \bigg(\sqrt{\frac{2}{\sqrt{1+4r^2}-1}}\bigg)$$
This happens specifically with differentiable complex functions. I have a kind of intuitive explanation of this phenomenon.
First, as I hope you know, every complex number can be written in polar form, which looks like this
$$r(\cos(\theta) + i\sin(\theta))$$
for some real $r \ge 0$ and $\theta \in \Bbb{R}$. Multiplying complex numbers in polar form is nice and intuitive. It's easy to verify that:
\begin{align*}
&r_1(\cos(\theta_1) + i\sin(\theta_1)) \times r_2(\cos(\theta_2) + i\sin(\theta_2)) \\
= \, &r_1 r_2(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)).
\end{align*}
What this means is, if we are multiplying a complex number $z$ in general by a specific complex number $r(\cos(\theta) + i\sin(\theta))$, then all we have to do is:
- Scale $z$ by a factor of $r$, and
- Rotate $z$ (counter-clockwise) by $\theta$ radians.
Note that both of these operations preserve lines and don't change angles. If we start with two lines in the complex plane intersecting at an angle $\alpha$, then multiplying every complex number in these lines by $r(\cos(\theta) + i\sin(\theta))$ will result in two (rotated) lines, still intersecting at an angle of $\alpha$.
Now, let's say we have a complex function $f$ which is differentiable at some complex number $z_0$. Essentially what this means is that the function $f$ behaves like its linearisation:
$$L(z) = f'(z_0)z + f(z_0)$$
for $z$ close to $z_0$. All that $L(z)$ does is multiply $z$ by a fixed complex number $f'(z_0)$ (which just scales and rotates), then adds a fixed complex number $f(z_0)$, which just shifts the picture without changing angles. So, $L$ will also not affect angles!
And since $f(z) \approx L(z)$ for $z$ around $z_0$, this means that, if two smooth curves meet at $z_0$ and their tangent lines meet at an angle $\alpha$, then whatever curves $f$ maps them to will have their tangents meet at angle $\alpha$.
So, what about your particular question? You have a complex function $f(a + ib) = u(a, b) + iv(a, b)$, which we will want to be differentiable. You are looking at the curves $u(a, b) = c_1$ and $v(a, b) = c_2$. The curve $u(a, b) = c_1$ is the set of points which will map to the vertical line $\operatorname{Re} z = c_1$. Similarly, $v(a, b) = c_2$ is the curve that will map to the horizontal line $\operatorname{Im} z = c_2$.
So, if the two curves intersect at $z_0$, and their tangents form an angle $\alpha$, then because $f$ is differentiable at $z_0$ (as it is everywhere else), the image of these curves must also form an angle $\alpha$. But, as we discussed, the image of these curves are horizontal and vertical lines, which are perpendicular! So, we must have $\alpha = \pi/2$. That is, if we look very closely at intersections between the curves, they must occur at right-angles.
Best Answer
It is common to misunderstand "$u$ is a function of $x$ and $y$" as "$u$ depends on $x$ and $y$". Which leads to errors such as thinking that $u(x,y)=x^2$ is not a function of $x $ and $y$.
Saying that "$u$ is a function of $x$ and $y$" means that the domain of $u$ is the set of ordered pairs $(x,y)$. Informally, $u$ takes two real numbers as inputs, and produces a number as output. To qualify as a function, $u$ must be consistent: if given the same pair $(x,y)$ today and tomorrow, it must give the same output. Other than that, it's free to use the inputs in any way it wants; and that includes not using them at all. For example, $f(x,y)=\sqrt{2}$ is a function which always returns $\sqrt{2}$ as output. It's still a function of $x$ and $y$ -- meaning it takes $x$ and $y$ as inputs -- it just does not use those inputs.