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I need some help in determining where $\log(z+z^{-1} -2)$ is analytic, where $z$ is a complex number and $\log(z)=\ln|z|+\arg(z+2k\pi),k\in\mathbb{Z}$. Thank you in advanced.

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  • $\begingroup$ You need to have the expression in parentheses fall inside of the region of analyticity. Logz is analytic in $\mathbb R^2 -[0,\infty)$, and the expression in the parentheses is analytic everywhere outside 0, since it is a rational function. In general, if you have g(f(z)), you want f(z) to fall in the region where g(z) is analytic. $\endgroup$ – DBFdalwayse Sep 1 '13 at 9:36
  • $\begingroup$ Note: The branch is the usual branch cut. $\endgroup$ – Gustavo Louis G. Montańo Sep 1 '13 at 13:55
  • $\begingroup$ You need to find the branch points first. See here. $\endgroup$ – Mhenni Benghorbal Sep 6 '13 at 5:15
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$$z+z^{-1}-2=\frac{z^2-2z+1}z=\frac{(z-1)^2}z$$

If you want to "delete" the usual branch cut for the logarithmic function, i.e. the ray $\,(-\infty,0]\;$ on the real axis, then it must be

$$\frac{(z-1)^2}z\notin (-\infty,0]\iff \text{Re}\left(\frac{(z-1)^2}z\right)>0\;\;\vee\;\;\text{Im}\left(\frac{(z-1)^2}z\right)\ne k\pi\;,\;k\in\Bbb Z$$

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    $\begingroup$ I think you meant something else regarding the imaginary part $\endgroup$ – lcv Feb 7 '14 at 10:27
  • $\begingroup$ Like I remember, @lcv... $\endgroup$ – DonAntonio Feb 7 '14 at 11:47
  • $\begingroup$ Maybe there is something I don't get but why your logarithm has a branch cut when the argument's imaginary part is an integer multiple of $\pi$? user64494's answer looks right to me. $\endgroup$ – lcv Feb 7 '14 at 23:04
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Too long for a comment. The Maple 17 commands $$expression := log(z+1/z-2); FunctionAdvisor(branch\_cuts, expression, plot = `2D`) $$ produce

enter image description here

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  • $\begingroup$ Maple 17 also describes the above set by a formula. Here is a part: $$\left( {\it Re} \left( z \right) =\sqrt {- \left( {\it Im} \left( z \right) \right) ^{2}+1},-1\leq {\it Im} \left( z \right) ,{\it Im} \left( z \right) <0 \right) ,\dots $$ $\endgroup$ – user64494 Sep 1 '13 at 11:22
  • $\begingroup$ 5-4 vote in favor of deletion. Not decisive enough for me. I will let other moderators to take a look, too, but I think this is, in some sense, the most useful of the three available ones :-) $\endgroup$ – Jyrki Lahtonen Sep 20 '15 at 8:06
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We have a composition f(g(z)) of functions, with $f(z)=Logz$, and $g(z)=\frac{(z-1)^2}{z}$ so we need g(z) to be defined in the domain of $Logz$, which is $\mathbb C-[0,\infty)$.

Let $w:=\frac{(z-1)^2}{z}$ .We want to exclude $w:=\frac{(z-1)^2}{z}$ from the branch cut $[0,\infty)$. This means we want to exclude the cases where:

1)$Im(w)=0$ and

2)$Re(w) \leq 0$

Notice we use and (instead of or) since we want to exclude points who have both a real part less than $0$ , and an imaginary part equal to $0$.

Expanding on w, letting $z:=(a+ib)$, we get :

3)$w= a^3-2a^2+a-2b^2+ab^2+i(a^2b+b^3-b)$ (I double-checked with Wolfram.)

Notice there is a nice factorization for the imaginary part of $w$;

4)$i(a^2b+b^3-b)=ib(a^2+b^2-1)$.

So, let's check #4 to see where 1),2) are satisfied:

From #4, we see that $Re(w)<0$ if either:

$I)b=0$

$II)a^2+b^2-1=0$

Let's check the case $I$ first , i.e., let's see what happens when $b=0$. If $b=0$,

using the expression for $w$ in #3 , we want to see where we get $b=0$ and $Re(w)<0$;

if $b=0$ , then$Re(w)=a^3-2a^2+a$. We see that $Rew=a(a-1)^2 \leq 0$ only when $a<0$, since

$(a-1)^2$ is always nonnegative. So we have that $Rew \leq 0$ precisely in $(- \infty,0]$

So we need to exclude the set :$Imz=0, Rez \leq 0$ , i.e., we need to exclude the negative real axis.

Now we must do something similar for the case where $a^2+b^2-1=0$ . Notice there is

a nice factorization for $Rew$; $Rew=(a^2+b^2)(a-2)+a$ . This should help you figure out

what part of $Rez$ to exclude for this case where $a^2+b^2-1=0$ . Can you finish it?

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