complex numberscomplex-analysislogarithmsreal numbers
Show that $\log\log z$ is analytic in the domain consisting of the $z$ plane with a branch cut along the line $y = 0$, $x \le 1$.
As of now I'm not too sure on how to solve this problem, so I was thinking you may have to use the Cauchy-Riemann equations to find the answer. I honestly tried it but I don't know what to do.
If someone can help me out in solving this problem that would be great. Thanks!
The way we define the Logarithm is the inverse of the exponential.
If $z= re^{i\theta}$,then $ln(z)=ln(r)+i\theta$.
Now , a problem that is there is, with this equation alone , logarithm is multi-valued i.e for every z there are infinitely many values for the logarithm . For $\theta$ differing by a value of $2\pi$ , one will get the same value. It isn't injective. This leads us restricting its domain so that each point(z) corresponds to only one value. This is referred to as a branch cut. Check the image from Wiki,
So, for log(z), simply remove any ray joining 0 and infinity.(Restricting complete rotation(0 to $2\pi$!). Also, see in the graph what happens. The principle log removes one particular ray, the negative real axis, I guess. This is only a convention, I guess.
Now, coming back to your question, you want log(iz)(PS:-The $iz$ only changes the labels on the axes) to be analytic in some given region. All you need to do is to remove a ray(not in the given region) to make it analytic in the given region.
$z=\pm2i$ are branch points from which branch cuts of order $2$ "emerge". For part (a), the branch cut must be chosen to be the straight line joining $z=-2i$ to $z=2i$. For part (b), the relevant branch cut has two segments: (i) one is the semi-infinite straight line from $z=2i$ to $z=i\infty$; and (ii)the other is the semi-infinite straight line from $z=-2i$ to $z=-i\infty$. The rest of the answer follows, doesn't it?
Best Answer
I think you have you to use the principal value of the logarithm. $\log z = \log |z| + \mathrm i\arg z$.