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desaila
Oct23-07, 10:49 PM
1. The problem statement, all variables and given/known data
f:Complex Plane ->Complex Plane by f(z) = (e^z - z^e)/(z^3-1) continuous? (Hint: it
has more than one discontinuity.)






3. The attempt at a solution

My attempt at a solution was thus, initially I expanded z^3 and tried to find where it equaled 1. That wasn't very straightforward, and then I figured I could just say the function is discontinuous when z^3 = 1, but I really don't think that's sufficient. Is there something I'm missing here?

Thanks in advance.

Dick
Oct23-07, 10:53 PM
z^e is not terribly well defined either. It's e^(log(z)*e) and log(z) has branches.

desaila
Oct23-07, 10:56 PM
z^e is not terribly well defined either. It's e^(log(z)*e) and log(z) has branches.

Well, if z^e isn't continuous is that really a problem? I thought that things in the numerator weren't really an issue when dealing with continuity.

jhicks
Oct23-07, 11:07 PM
Are you familiar with the nth "roots of unity"?

Dick
Oct23-07, 11:08 PM
Well, if z^e isn't continuous is that really a problem? I thought that things in the numerator weren't really an issue when dealing with continuity.

Not true. A discontinuous numerator is just as bad as a zero in the denominator.

desaila
Oct23-07, 11:12 PM
Are you familiar with the nth "roots of unity"?

No. I'll wiki this, but I'd be interested in your explanation as well.

Not true. A discontinuous numerator is just as bad as a zero in the denominator.

Really. So z^3 = 1 wouldn't be so bad, I'd just have to find discontinuities in e^z and z^e then?

EDIT: After reading about the nth roots of unity I could provide them for z^3 = 1. They're given on the wiki entry( http://en.wikipedia.org/wiki/Roots_of_unity )

Dick
Oct23-07, 11:18 PM
No, z^3=1 is bad too. That's three discontinuous points. e^z doesn't have any. log(z) has a whole line of discontinuities. So that goes for z^e as well.

desaila
Oct23-07, 11:19 PM
No, z^3=1 is bad too. That's three discontinuous points. e^z doesn't have any. log(z) has a whole line of discontinuities. So that goes for z^e as well.

What I meant was showing the discontinuities of z^3 = 1 wouldn't be too bad. Sorry about the misunderstanding.

desaila
Oct23-07, 11:26 PM
Also, Dick, is there somewhere I can read up on z^e? Why it's defined the way you say it is, etc.

Dick
Oct23-07, 11:35 PM
There's no other way to define it. z=e^(log(z)) so z^e=e^(log(z)*e). But there is no continuous definition of log(z) in the whole complex plane. You have to remove part of the domain to get a continuous function. Look up "branch cuts". It's not just z^e. sqrt(z) has a very similar problem.

HallsofIvy
Oct24-07, 07:12 AM
What exactly is the question? It appears to ask whether or not the function is continuous but then immediately tells you that it is not! Is the problem to find the points of discontinuity?

SiddharthM
Oct24-07, 07:12 AM
generally the numerator also matters..for example. Sinz/z doesn't REALLY have a discontinuity at the origin (i.e the origin is NOT a pole) whereas 1/z DOES have a legitimate discontinuity at the origin(it's a simple pole).

you need to understand the concept of branch cuts. also computing complex roots (wen the numbers are simple of course) is straightforward and can and probably should be done by hand.