i^i and other complex powers -- Complex Analysis 5

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Күн бұрын

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@Zeitgeist9000 2 жыл бұрын

Man this series is several orders of magnitude better than the complex analysis course I had in college many moons ago. Kinda makes me mad, but I'm grateful to be seeing it presented so well.

@jokarmaths7771 2 жыл бұрын

amazing ............... kzbin.info/www/bejne/eXbdaaePmrWontU

@geekmath-ux7zj Жыл бұрын

My answers for these warm-up questions: Q1 (1) e^(pi/2+2*pi*n) (2) i*e^-(pi/2+2*pi*n) (3) e^-(pi/4+2*pi*n)*e^(i*0.5*ln(2)) Q2 e^(i*(alpha+beta)*2*pi) Q3 (1)the phase factor in (-1,0)U(1,infinity) is -1 (2)in the complex plane. Thank you professor for your wonderful explanation.

@deltalima6703 2 жыл бұрын

I like these complex videos. They are interesting. I looked at other ones and they dont usually do things rigorously enough and it just seems suspicious. These are perfect. Imho.

@jokarmaths7771 2 жыл бұрын

amazing ............... kzbin.info/www/bejne/eXbdaaePmrWontU

@howardcheung8304 2 жыл бұрын

Well explained!

@jokarmaths7771 2 жыл бұрын

amazing ............... kzbin.info/www/bejne/eXbdaaePmrWontU

@Grecks75 Ай бұрын

For the complex power function w=z^a (for complex z,a), the branch cut is usually made along the *non-positive* real axis (the same as with complex logarithm), in the texts and presentations I have seen. So the question arises: Is there any *standard* way at all to define the branch cut and the principal value of the complex power function?

@nathanisbored 2 жыл бұрын

the example at 15:39 kinda lost me... i dont know what this has to do with the Lemma on the previous board. it doesnt seem to fit the form at all, and if i chose a g(z) to make it fit, the g(z) wont be single-valued. furthermore im not sure why we decided to split the function up in the way we did, or why its allowed to decompose and study phase factors separate like that. also, when you discuss the sqrt(1-z) part, you seem to imply the relevant z_0 point is at 1, which means from the Lemma it should be sqrt(z - 1), not sqrt (1 - z). im also curious what the formal definition of a phase factor is... presumably some limit thing.

@ConManAU 2 жыл бұрын

First, notice that sqrt(1-z) still takes the value sqrt(0) when z=1, and that’s what identifies the origin point for the branch cut and phase factors. You could also think of it as being i sqrt(z-1), if you want to explicitly write it in the form of the lemma. As for the way the function is broken up, it was maybe not clear in the video but if you avoid going around the problem point z_0 then the function essentially stays single-valued. So, for example, if you consider a circle around z=1 of radius less than 1, the sqrt(z) term behaves nicely and you can treat it as the f(z) from the lemma, and similarly for the sqrt(1-z) term in the vicinity of z=0. Like Michael said, you can put the branch cuts going in any direction from the source, so the branch cut for sqrt(z) can be along any ray from the origin (i.e. of the form Arg z = t for some t), and similarly for the sqrt(1-z) term coming from z=1. As it happens, in this case since the phase factors cancel out, if you make the branch cuts going right from z=0 and left from z=1, they cancel out, but if they compounded (for example, maybe sqrt(z/(z-1))?) then you’d just have a variety of branch cuts with different phase factors. Could this have been explained better in the video? Probably, but it would have made the video much longer than it already is.

@jokarmaths7771 2 жыл бұрын

amazing ............... kzbin.info/www/bejne/eXbdaaePmrWontU

@sillymel 2 жыл бұрын

@nathanisbored Wow. Didn’t expect to see you here.

@Grecks75 Ай бұрын

I agree very much. That was very sloppy and I have similar doubts. First, when you treat the complex sqrt() function as single-valued (taking its principle value), splitting up the square root does NOT work in general. Second, I don't see why the choice of a specific branch cut for the sqrt function should be arbitrary, because it directly affects its definition, namely the very definition of what its principal value is. Changing the branch cut changes the principal value and thus the definition of the (single-valued) function. Third, I think the way the branch cut for sqrt(1 - z) was initially drawn was wrong: If you decide for the sqrt(z) function to have a branch cut extending along the Arg z = 0 ray, then the second sqrt part, function h(z) = sqrt(1 - z), should have a branch cut starting at 1 and extending in the opposite direction, i.e. in the Arg z = pi direction.

@SzanyiAtti 2 жыл бұрын

9:52 I thought we used the convention that Arg(z) is on the interval (-pi, pi], so wouldn't theta approach 0 from the left? Or did we switch to the convention that Arg(z) is on the interval [0, 2pi)? What am I missing?

@schweinmachtbree1013 2 жыл бұрын

yes I think it's just that the convention is different - using the (-pi, pi] convention the branch cut would just be on the other side (so one would consider approaching a point on the negative real axis instead of the positive real axis).

@jokarmaths7771 2 жыл бұрын

amazing ............... kzbin.info/www/bejne/eXbdaaePmrWontU

@abusayeed3613 2 жыл бұрын

Make a series on numerical analysis

@yaroslavshustrov2823 2 жыл бұрын

Is there any place where I could find solutions to the problems at the end of the videos? Or at least the answers, so I could check myself. Thanks in advance.