Question

log z= log|z| + iarg z , \(\alpha\leq arg z\leq \alpha + 2\pi\)
Hence logz is analytic on \(\mathbb C -\{re^{i\alpha}\}, r\geq 0\)
My question: what is alpha? how to find it? Please, help

Answer 1

@oldrin.bataku

Answer 2

the point is that you can choose any \(\alpha\), and you'll get a valid branch of \(\log z\) with a branch cut along \(re^{i\alpha},r\ge 0\)

Answer 3

Give me a concrete example, please.

Answer 4

if you choose the range \([\alpha,\alpha+2\pi)\) for \(\arg z\) then we get an analytic branch of \(\log z\) aside from the branch cut along the ray \(z=re^{i\alpha}\)

Answer 5

the standard example is with \(\alpha=-\pi\), so we get arguments in \([-\pi,\pi)\) and a branch cut along the ray \(re^{-i\pi}=-r,r\ge 0\), aka the negative real axis \((-\infty,0)\)

Answer 6

Now, let \(z =\sqrt2 /2 + i\sqrt 2/2 \) hence arg z = pi/4,
How can I define alpha?

Answer 7

but we could just as easily decide we want angles between \(0,2\pi\) so \(\alpha=0\), though this gives us instead a branch cut on the positive real axis \((0,\infty)\) since \(re^{i0}=r,r\ge 0\) which is somewhat less convenient usually as we end up with no analyticity on the reals

Answer 8

Should I take alpha = pi/4 also?
hence the region is \(C - \{re^{i\pi/4}\} , r\geq 0\)

Answer 9

what do you mean take \(z=e^{i\pi/4}\), you need to pick a range of angles to use for the plane

Answer 10

I didn't say that. I said if \(z = \sqrt 2/2 + i\sqrt 2/2\) then \(arg z = \pi/4\)

Answer 11

that's what branch cuts have to do with; we have a Riemann surface that covers the plane however many times, and then we take a 'section' of it and deal with the 'defect' we get along the borders

Answer 12

that doesn't narrow it down, @Loser66 -- that is true in any interval that contains \(\pi/4\), including: $$(-\pi,\pi),\quad(0,2\pi),\quad(\pi/4,2\pi+\pi/4),\text{ etc.}$$

Answer 13

Is there any video clip I can study ? I want to completely understand what it is. Please.

Answer 14

the point is that if $$z=e^{i\theta}$$ then it follows \(z=e^{i(\theta+2\pi k)}\) for all \(k\in\mathbb Z\), so we need to fix some nonambiguous set of angles to use for \(\arg z\)

Answer 15

the way to make \(\log z\) close to analytic is to pick some contiguous interval like \([\alpha,\alpha+2\pi)\), which works well, except until we get close to an argument of \(\alpha+2\pi\) and then it suddenly 'jumps' back down to \(\alpha\), wrapping around

Answer 16

this is the 'gap' you see between the different 'leaves' or branches of the complex logarithm here:

https://upload.wikimedia.org/wikipedia/commons/thumb/a/ab/Riemann_surface_log.svg/791px-Riemann_surface_log.svg.png

Answer 17

that's the reason analyticity fails along the ray \(\{re^{i\alpha}:r\ge 0\}\) -- this is the ray over which the argument suddenly jumps by \(2\pi\) in a discontinuous way: