# Find the range and inverse of $\log(z^2+1)$.

This was a three part problem. In the first part I showed that for $A=\{z:\Re(z)>0\}$ that the function $f:A\rightarrow\mathbb{C}$ given by $f(z)=\mathrm{Log}(z^2+1)$ was univalent. (Log here is defined as the principal log).

Now I am trying to find the range of $f$ and its inverse $f^{-1}$.

For the range, I broke $f$ down into its components $g(z)=z^2$, $h(z)=z+1$ and $l(z)=\mathrm{Log}(z)$. For the function $g$, I saw that for any $z\in A$, if $z=x+iy$ and $x>0$ then $z^2=(x^2-y^2)+i2xy$ where $(x^2-y^2)$ and $2xy$ can be any number in $\mathbb{R}$ so $g:A\rightarrow\mathbb{C}$. For $h\circ g=z^2+1$ I thought that since $z^2\in\mathbb{C}$ then it followed that $z^2+1\in\mathbb{C}+1$ so $h\circ g:A\rightarrow\mathbb{C}+1$. Finally, since $\mathbb{C}+1\subseteq\mathbb{C}$ we have $\mathrm{Log}(z^2+1)$ unrestricted on the real axis but since it is the principal log it is bounded on the imaginary axis in the interval $(-\pi,\pi]$.

For the inverse function I let $w=\mathrm{Log}(z^2+1)$. Then

$$e^w=z^2+1\Leftrightarrow e^w-1=z^2\Leftrightarrow \sqrt{e^w-1}=z$$

where

$$\sqrt{e^w-1}=\sqrt{|e^w-1|}\left[\cos\left(\frac{\arg(e^w-1)}{2}\right)+i\sin\left(\frac{\arg(e^w-1)}{2}\right)\right]$$

if $\sqrt{|e^w-1|}\cos\left(\frac{\arg(e^w-1)}{2}\right)>0$ or

$$\sqrt{e^w-1}=\sqrt{|e^w-1|}\left[\cos\left(\frac{\arg(e^w-1)}{2}+\pi\right)+i\sin\left(\frac{\arg(e^w-1)}{2}+\pi\right)\right]$$

if $\sqrt{|e^w-1|}\cos\left(\frac{\arg(e^w-1)}{2}+\pi\right)>0$.

I am particularly worried about the range of $\mathrm{Log}(z^2+1)$ because for the range that I found, we could possibly have $w=0$ which would present a problem for the inverse function that I found where $z$ cannot be equal to 0.

• Actually $h\circ g(A) = \{z\in\mathbb C : \Re z > 0 \vee \Im z \neq 1\}$. You can remove $\{z\in\mathbb C : \Re z = 0 \wedge \Im z \in [0,1] \} = "[0,i]"$ from this to get a "slash" from $0$ to $\infty$ and thus to be able to invert $\log(z^2+1)$ in that region. – AlexR Jan 20 '14 at 9:28
• Why is $\Re(z)>0$ for the range of $h\circ g(A)$? Couldn't $x^2-y^2$ be any value in $\mathbb{R}$? – A. Wong Jan 20 '14 at 9:31
• Note the $\vee$. It says, if $\Im z = 1$ then $\Re z > 0$ or it's not in the image. – AlexR Jan 20 '14 at 9:32
• You are omitting a branch of the square root: $e^w-1=z^2\Leftrightarrow z = \pm \sqrt{e^w-1}$ – gammatester Jan 20 '14 at 9:33
• @AlexR I am a little confused with the notation. I am reading it as $z\in\mathbb{C}$ such that $\Re(z)>0$ or $\Im(z)\not=1$. – A. Wong Jan 20 '14 at 9:38

$z^2$ maps $A$ onto $\mathbb C \setminus (-\infty,0]$

here you missed that if $a=x^2-y^2$ and $b=2xy=0$ then $y=0$ and therefore $a>0$

$z^2+1$ maps $A$ onto $\mathbb C \setminus (-\infty,1]$

Since $\operatorname{Log}$ is a biholomorphic map $$\operatorname{Log}: \mathbb C \setminus (-\infty,0] \rightarrow \mathbb \{z\in \mathbb C : -\pi < \operatorname{Im}z< \pi \}$$ you see that the range of $\operatorname{Log(z^2+1)}$ is the set $$\{z\in \mathbb C : -\pi < \operatorname{Im}z< \pi \} \setminus \operatorname{Log}((0,1])= \{z\in \mathbb C : -\pi < \operatorname{Im}z< \pi \} \setminus (-\infty,0]$$

• What if $y>x$? Couldn't we then have $a<0$? – A. Wong Jan 20 '14 at 9:46
• Yes, any $x+iy$ having $y>x>0$ is in $z^2(A)$, but in that case $2xy>0$ – Blah Jan 20 '14 at 9:59
• You can solve $x^2-y^2=a$, $2xy=b$ always with a pair $(x,y)$ such that $x>0$ if $b\neq 0$! – Blah Jan 20 '14 at 10:02
• Then wouldn't $z^2$ maps A onto $\mathbb{C}\backslash(-\infty,0]$ be excluding that? – A. Wong Jan 20 '14 at 10:03
• Just show that $F(x,y):A \rightarrow \mathbb C \setminus (-\infty,0]$ given by $F(x,y)=(x^2-y^2,2xy)$ is injective and surjective... – Blah Jan 20 '14 at 10:07

You might find this picture to be illustrative:

The picture maps the positive real half-disk $\{re^{i\theta} : r > 0, -\frac{\pi}{2} < \theta < \frac{\pi}{2} \}$ to $\mathbb{C} \smallsetminus (-\infty, 0]$. Lines indicate curves of constant $|z|$ or $\arg(z)$. Hue is argument; saturation and brightness indicate magnitude.

The Mathematica code used to generate this picture is as follows:

ParametricPlot[{Re[#], Im[#]} &[Log[(r Exp[I t])^2 + 1]],
{r, .001, 5}, {t, -Pi/2, Pi/2},
PlotPoints -> 50, MaxRecursion -> 3, PlotRange -> {{-3, 3}, {-3, 3}},
Mesh -> {30, 30}, ColorFunction -> Function[{x, y, u, v}, Hue[v,
1/(0.25 + 3 Log[u + 1]), 1 - 1/(1.1 + 10 Log[u + 1])]]]