# Multifunctions

• Apr 26th 2011, 03:23 AM
Alexrey
Multifunctions
Hey guys,

I'm having some serious trouble with the concept of multifunctions. My lecturer went over them briefly and set a few questions in our tutorial but I'm not exactly sure what he wants us to do. For example, find:

[[i^1/6]]

What exactly does he want us to find? Is this question saying "find the multifuncitons of i^1/6? I spent literally a whole day trying to understand multifunctions by reading my lecture notes, going online and reading T. Needhams, Visual Complex Analysis chapter on the subject but have not really made that much progress on the matter. Am I correct in saying that a multifunction is a "function" that has multiple values for the same value of a complex number assigned to it? If so then I think I understand why log(z) is a multifunction, and I'm beginning to see why fractional powers of z are also multifunctions, but surely then {z}^{n} is also a multifunction? We did a section on finding a complex number's "roots of unity", with {z}^{n} having exactly n roots. But like I said, aren't these roots just multifunctions of z?

Any help would be appreciated. I'm really struggling with Complex Analysis as a whole, so getting rid of one of my many problems of understanding in this subject would really lower my stress levels. I apologise for not being able to use LATEX correctly.

Thanks guys, cheers.
• Apr 26th 2011, 04:32 AM
Opalg
Quote:

Originally Posted by Alexrey
I'm having some serious trouble with the concept of multifunctions. My lecturer went over them briefly and set a few questions in our tutorial but I'm not exactly sure what he wants us to do. For example, find:

[[i^1/6]]

What exactly does he want us to find? Is this question saying "find the multifuncitons of i^1/6? I spent literally a whole day trying to understand multifunctions by reading my lecture notes, going online and reading T. Needhams, Visual Complex Analysis chapter on the subject but have not really made that much progress on the matter. Am I correct in saying that a multifunction is a "function" that has multiple values for the same value of a complex number assigned to it? If so then I think I understand why log(z) is a multifunction, and I'm beginning to see why fractional powers of z are also multifunctions, but surely then {z}^{n} is also a multifunction? We did a section on finding a complex number's "roots of unity", with {z}^{n} having exactly n roots. But like I said, aren't these roots just multifunctions of z?

The expression z^{1/6} represents a multifunction, in the sense that every complex number has six 1/6-th roots. All that this question is asking you to do is to use de Moivre's theorem to find the six values of i^{1/6}.

Quote:

Originally Posted by Alexrey
I apologise for not being able to use LATEX correctly.

Not your fault. This site is currently having serious problems with its LaTeX engine.
• Apr 26th 2011, 04:57 AM
topsquark
Quote:

Originally Posted by Alexrey
I apologise for not being able to use LATEX correctly.

See post 8 in this thread for a temporary solution.

-Dan
• Apr 26th 2011, 05:15 AM
Alexrey
Thanks for the help guys. I had a look at the solution to that particular question and it seemed exactly the same as finding the roots of unity for z^6 = i which was just e^i(pi/12 + k pi/3) for k={0,1,...,5}, so doesn't that mean that z^n is also a multifunction, or are multifunctions just restricted to z^1/n and log(z)?
• Apr 26th 2011, 07:20 AM
Opalg
Quote:

Originally Posted by Alexrey
Thanks for the help guys. I had a look at the solution to that particular question and it seemed exactly the same as finding the roots of unity for z^6 = i which was just e^i(pi/12 + k pi/3) for k={0,1,...,5}, so doesn't that mean that z^n is also a multifunction, or are multifunctions just restricted to z^1/n and log(z)?

z^r is an honest (single-valued) function if r is an integer. But if r is fractional then z^r is a multifunction, having as many possible values as the denominator of the fraction.

I think that you may be confusing the function z^r with the equation z^r = k. The function z^r is a multifunction whenever r is not an integer. But the equation z^r = k has r solutions when r is an integer. That is because it is equivalent to the equation z = k^{1/r}. It is the fractional power 1/r that leads to the multiplicity of values.
• Apr 26th 2011, 09:29 AM
Alexrey
Wow, okay that went totally over my head. I clearly need to read over the multifunctions section some more, but no matter how many times I do it I still don't understand why z^1/n (n an integer) is a multifunction whereas z^n (n an integer) is not. Complex Analysis truly is giving me a hard time. Second year maths was so damn easy compared to third year :(. I've spent almost every waking minute since Friday morning doing CA work, but it still seems so exotic to me. Hopefully I'll have a lightbulb moment sometime soon.

Thanks very much for your help Opalg.

By the way, are these analysis courses (complex and real analysis) all based around being able to prove theorems and statements? I looked at some past exam papers and they all seemed to say prove this or prove that, with hardly any calculation work involved as I've been used to for the past 2 years.