I want to describe the mapping of the domain $D =\{z \in \mathbb{C}: \text{Re}(z) >0\}$ under the function $f(z) =\sqrt{z^2+1}$. (Using the Principal branch of the logarithm to denote the complex square root). I have tried several things, but every time I failed to come to a good description of this mapping.
First, I thought it would be useful to try and describe $f(x,y) = u(x,y) + i\cdot v(x,y).$ With $u$ and $v$ being real-valued functions of $x$ and $y$. (We're also assuming here that $z=x+i \cdot y $) However, when trying this, I find a very difficult function for $f(x,y)$, therefore I can't really say anything about the mapping.
I found the following function for $f(x,y)$: $$ f(x,y) =\exp \{ \frac{1}{2}( \ln(|x^2+y^2-1+i\cdot2xy|)+i\cdot \text{Arg}(x^2+1-y^2+i\cdot2xy))\}. $$
By using the fact that any $z\in \mathbb{C}$ can be written as $z=re^{i \theta}$, we dont get an easier expression for $f(r,\theta)$.
Then I thought perhaps it would be a better approach if we were to first find the mapping of $h(z) =z^2+1$ and then map the mapping of $h(z)$ using $g(z) = \sqrt{z}$.
So first I tried writing $h(x,y) = u_h(x,y)+i \cdot v_h(x,y)$, in which I succeeded. I found the following:
$$ h(x,y) = x^2-y^2+1+i\cdot 2xy. $$ It's easy to find the mapping of D under $h(x,y)$, by first appointing a value $c > 0$ to $x$ and then letting $y$ be any value in $\mathbb{R}$. We do the same with taking $ d \in \mathbb{R}$ for $y$ and then letting $x$ be any value bigger than 0. We then get horizontal parabolas.
Then, we need to find the mapping of these horizontal parabolas under $z \mapsto \sqrt{z}$. However, then we encounter a similar problem to what I found in the first method, using this way of trying to find a parametric curve in $\mathbb{R}^2$ that corresponds with our complex mapping, is rather difficult for the squareroot function. I tried and found the following:
$$ g(x,y) = \ln|x+i\cdot y|\cos(\frac{1}{2}\text{Arg}(x+i\cdot y))+ i \cdot \ln|x+i\cdot y|\sin(\frac{1}{2}\text{Arg}(x+i\cdot y)). $$
But I wouldn't know what the mapping of the horizontal parabolas would look like under this mapping.
So my question is: How can I describe the mapping of this function? Is there a general way of doing this for any function?
Firstly, look at the imaginary axis $x=0$. It is the boundary of the region.
Mostly, $yi$ ends up at $i\sqrt{y^2-1}$.
Then, for large $z$, we have $$\sqrt{z^2+1}=z\sqrt{1+\frac1{z^2}}\\ \approx z(1+\frac1{2z^2}-\cdots)\\ \approx z+\frac1{2z}-\cdots$$ So a long way from the origin, the grid lines don't move much.