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?

  • $\begingroup$ Maybe you find this helpful: bit.ly/2KOHCqe $\endgroup$ – James Jul 4 '18 at 13:22
  • $\begingroup$ @Lubin What I have learned to do, is to take horizontal and vertical lines in the domain and look at how the mapping changes these in the codomain. I'm not sure how to utilize the hints you've given me. $\endgroup$ – K.Kamal Jul 4 '18 at 19:31
  • $\begingroup$ If you let $z=t+\alpha i$, so running along a horizontal line, and look at the corresponding values of $\sqrt{1+z^2}$, you seem to get a quartic curve that goes out to infinity along asymptotes of slope $\pm1$. (Maybe) $\endgroup$ – Lubin Jul 4 '18 at 19:32

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.


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