Complex Analysis: Sketch the image of a domain under the mapping $f(z)=\sqrt{z^2+1}$

Question

I'm having some trouble solving the following:

Suppose $D=\{z: Re(z)>0\}$ and suppose $f(z)=\sqrt{z^2+1}$ where we mean the principal branch of the squared root. Find and sketch the image of $D$ under $f$.

I've tried to rewrite $f(z)=w=u+iv$ in terms of $x$ and $y$ to determine a $u$- and $v$-part which I could try to sketch. So I got this:

$w=\sqrt{z^2+1}=\sqrt{(x+iy)^2+1}=\sqrt{x^2+i2xy-y^2+1}$.

But this is where I'm stuck. I can't seem to get rid of the squared root. Also, I have yet to see how to proceed after rewriting. Can anyone help me out?

Answer 1

$z \to z^2 +1$ maps the right half-plane $D$ onto $\Bbb C \setminus (-\infty, 1]$ =: G, so it remains to determine the image of $G$ under the principal branch of the square root.

Now note that the principal branch of the square root is a one-to-one map of $\Bbb C \setminus (-\infty, 0]$ onto the right half-plane $D$, and maps the interval $(0, 1]$ onto itself.

It follows that $f(D) = D \setminus (0, 1]$.

Answer 2

The square root of a complex number can be found as follows:

$$\sqrt{x+iy}=u+iv\implies x+iy=u^2-v^2+i2uv$$ and you need to solve the system

$$\begin{cases}u^2-v^2=x,\\2uv=y.\end{cases}$$

Multiplying the first equation by $u^2$,

$$u^4-xu^2-\frac{y^2}4=0$$ gives the solution

$$u^2=\frac{x\pm\sqrt{x^2+y^2}}2,$$ then

$$u=\pm\sqrt{\frac{x+\sqrt{x^2+y^2}}2}$$ and $v$ follows.

To sketch the transformation, you can draw the iso-$x$ and iso-$y$.