# Given n points on the unit circle

ModernLogic
Let a0, . . . , an be points on the unit circle. Show that there is some other point p on the unit circle such that the product of the distances from p to ai for i=0,...,n at least 1. (Hint: Maximum Modulus Principle)

Maximum Modulus Principle: Let f be a nonconstant holomorphic function in the open connected subset G of C. Then absolute value of f does not attain a local maximum.

Man, I'm so stuck on this problem. I can't seem to figure out how the maximum modulus principle relates to the problem. Besides, the problem is asking me to find the lower bound for the product whereas the the maximum modulus principle states that there is no upper bound. I'm so confused.

Anything would help from you geniuses out there: A hint or advice.

Regards,
Steve

## Answers and Replies

matt grime
Homework Helper
That is one statement of the maximum modulus principle, which can be restated in terms of closed sets. in this case it could be taken to mean that any holomorphic function on a closed disc attains it maximum absolute value on the boundary.

What is the distance from p to the a_i? it is |p-a_i|, so we are looking at the product of these things so we want to consider the

f(x)=(x-a_0)(x-a_1)(x-a_2)..(x-a_n)

when x is inside (or on) the unit circle, now the maximum modulus principle tells us that |f(x)| is attained on the boundary, all we need do now is show that at some point in the interior |f(x)|=1 and we are done. obviously there is only one distinguished point in the interior of the disc....

ModernLogic
matt grime said:
That is one statement of the maximum modulus principle, which can be restated in terms of closed sets. in this case it could be taken to mean that any holomorphic function on a closed disc attains it maximum absolute value on the boundary.

Interesting. Thanks for the insight, Matt.

But just because a nonconstant holomorphic function doesn't attain a local maximum on an open disc, doesn't imply that it does on a closed disc. At least, I can't make that immediate connection.

matt grime
Homework Helper
the maximum modulus principle simply states that on a closed bound set such as the disc that the maximum occurs on the boudnary exactyl because there are no local maxima on the interior:

a holomorphic function is continuous, thus |f(x)| is continuousand hence on any compact set (closed and bounded) |f| is bounded and attains its bounds, it cannot happen on the interior by the maximum modulus principle so it happens on the boundary.

i am not claiming a local maximum (ie a turning point) on the boundary (it may not even be defined outside the boundary) but a global maxmium on the closed set.

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ModernLogic
matt grime said:
the maximum modulus principle simply states that on a closed bound set such as the disc that the maximum occurs on the boudnary exactyl because there are no local maxima on the interior:

a holomorphic function is continuous, thus |f(x)| is continuousand hence on any compact set (closed and bounded) |f| is bounded and attains its bounds, it cannot happen on the interior by the maximum modulus principle so it happens on the boundary.

i am not claiming a local maximum (ie a turning point) on the boundary (it may not even be defined outside the boundary) but a global maxmium on the closed set.

What about this though: Consider the holomorphic function f(x) = -(X-2)(X-3) which achieves a global maximum somewhere between 2 and 3. Now define this function on the interval [2,3]. |f(x)| doesn't achieve its maximum on the boundary.

Once again, thanks for your insights, Matt. They've been very help thus far.

matt grime
Homework Helper
ModernLogic said:
What about this though: Consider the holomorphic function f(x) = -(X-2)(X-3) which achieves a global maximum somewhere between 2 and 3.

Does it? I don't think it does, since we are talking about functions defined on the complex plane. And anyway, we are talking about the abs value anyway.

Now define this function on the interval [2,3]. |f(x)| doesn't achieve its maximum on the boundary/

but we aren't talking about f as a function on a subset of C with the boundary taken there, are we? Remember we are talking about discs in the complex plane and their boundaries.

ModernLogic
matt grime said:
Does it? I don't think it does, since we are talking about functions defined on the complex plane. And anyway, we are talking about the abs value anyway.

but we aren't talking about f as a function on a subset of C with the boundary taken there, are we? Remember we are talking about discs in the complex plane and their boundaries.

Oh, I see. They have to be discs. Cause [2,3] is still a subset of C.

matt grime