- #1
InbredDummy
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1) Let U be a subset of C s.t U is open and connected and let f bea holomorphic function on U s.t. for every z in U, |f(z)| = 1, ie takes takes all points in U to the boundary of the unit circle. Prove that f is constant.
Pf.
Suppose f is not constant. Then we can find a w s.t. f'(w) is nonzero, else we know that f is holomorphic and hence has a power series representation, therefore since every w has zero derivative, f = a0, the first coefficient in the power series representation, ie f is constant. So let w be a point where f has nonzero derivative. We can therefore rewrite f(z) = f(w) + f'(w)(z-w) + h(z)(z-w) where h(z) --> 0 as z --> w, (WHY?, i don't recall why this is so, it's just in my notes for some reason). So f is nonconstant and analytic, therefore f is an open mapping. so consider some open neighborhood of w, call this neighborhood V, we know that f takes V to another open set f(V).
Here is where i am stuck, what gurantees that this point is not in the open neighborhood intersected with the boundary of the circle? i know it has something to do with us rewriting f(z) in terms of f(w) and h(z).
2) I guess this is a similar question, Suppose that f and g are holomorphic in a connected open set U and |f(z)|^2 + |g(z)|^2 = 1, show that f and g are constant.
Pf.
I have that u |--> (f(u), g(u)) which sits in C^2.
So converting to real coordinates, i get that: f(u) = x1 + iy1, g(u) = x2+iy2
so (x1)^2 + (y1)^2 + (x2)^2 + (y2)^2 = 1, giving me a 3 dimensional sphere sitting in R^4.
Now suppose that f and g are not constant, again, since they are holomorphic they are also analytic and also there exists a w in U with nonzero derivative, else f, g are constant functions.
So let's take a w with nonzero derivative and consider an open neighborhood around w, now both f and g are again open mappings, so it takes our open neighborhood around w, let it be denoted by V, to another open neighborhood around (f(w), g(w)). But again, how do i derive that the neighborhood of points contains a point that is not on the boundary of our sphere?
3) Let f be an entire function, N a positive integer and C > 0
(a) Suppose that |f(z)| <= (less than or equal to) C|z|^N for every z in C. Show that f is a polynomial.
Pf.
I'm not very sure where to start on this one. I know that polynomials are entire functions. I'm guessing we want to show that if f has a power series representation, then after some positive integer G, for every n > G, an = 0?
(b) Suppose that |f(z)| => (greater than or equal to) C|z|^N for |z| large enough. Show that f is polynomial.
Pf.
eh not really sure again, sorry guys, I am kinda stumped on this one.
(4)
Consider the function f(z) = 1 + z^2 + ...+ z^(2^(n-1)) + ... and |z| < 1.
Show that for every a = e^(k*2pi*i/2^n), n greater than or equal to 0, k = 1, 2, ..., 2^n-1, that f(z) tends to infinity when z approaches a along the radius of the circle, lim (r>0, r--> 1 from the negative side) f(r*a) = infinity. Conclude that every point of the unit circle is a singularity for f.
Pf.
Hmm, well I first of all don't really even understand the question. We are saying, given an n and given an r that satisfies our limit condition then
f(z) --> infinity? and what exactly is a supposed to be?
(5) Suppose that f(z) is given by a power series and has radius of convergence equal to 1 and there are only poles of first order on the unit circle (no other singularities). Show that the sequence {an} is bounded.
Pf.
? lol
(6) Suppose we wanted to represent the function
f(z) = 1/(1-z^2) + 1/(3-z) by a Laurent series. How many such representations are there? In which region is each of them valid? Find the coefficients aj explicitly for each of these representations.
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I know that there is more than 1 singularity, but they are isolated singularities. and there are 3 ways to represent a function by a Laurent series, one is by the power series if the function is continuous, however this is not a continuous function, the 2nd is by by terms with negative indice coefficients, and the third is by an infinite series. Need some guidance on this one again.
Thank you!
Pf.
Suppose f is not constant. Then we can find a w s.t. f'(w) is nonzero, else we know that f is holomorphic and hence has a power series representation, therefore since every w has zero derivative, f = a0, the first coefficient in the power series representation, ie f is constant. So let w be a point where f has nonzero derivative. We can therefore rewrite f(z) = f(w) + f'(w)(z-w) + h(z)(z-w) where h(z) --> 0 as z --> w, (WHY?, i don't recall why this is so, it's just in my notes for some reason). So f is nonconstant and analytic, therefore f is an open mapping. so consider some open neighborhood of w, call this neighborhood V, we know that f takes V to another open set f(V).
Here is where i am stuck, what gurantees that this point is not in the open neighborhood intersected with the boundary of the circle? i know it has something to do with us rewriting f(z) in terms of f(w) and h(z).
2) I guess this is a similar question, Suppose that f and g are holomorphic in a connected open set U and |f(z)|^2 + |g(z)|^2 = 1, show that f and g are constant.
Pf.
I have that u |--> (f(u), g(u)) which sits in C^2.
So converting to real coordinates, i get that: f(u) = x1 + iy1, g(u) = x2+iy2
so (x1)^2 + (y1)^2 + (x2)^2 + (y2)^2 = 1, giving me a 3 dimensional sphere sitting in R^4.
Now suppose that f and g are not constant, again, since they are holomorphic they are also analytic and also there exists a w in U with nonzero derivative, else f, g are constant functions.
So let's take a w with nonzero derivative and consider an open neighborhood around w, now both f and g are again open mappings, so it takes our open neighborhood around w, let it be denoted by V, to another open neighborhood around (f(w), g(w)). But again, how do i derive that the neighborhood of points contains a point that is not on the boundary of our sphere?
3) Let f be an entire function, N a positive integer and C > 0
(a) Suppose that |f(z)| <= (less than or equal to) C|z|^N for every z in C. Show that f is a polynomial.
Pf.
I'm not very sure where to start on this one. I know that polynomials are entire functions. I'm guessing we want to show that if f has a power series representation, then after some positive integer G, for every n > G, an = 0?
(b) Suppose that |f(z)| => (greater than or equal to) C|z|^N for |z| large enough. Show that f is polynomial.
Pf.
eh not really sure again, sorry guys, I am kinda stumped on this one.
(4)
Consider the function f(z) = 1 + z^2 + ...+ z^(2^(n-1)) + ... and |z| < 1.
Show that for every a = e^(k*2pi*i/2^n), n greater than or equal to 0, k = 1, 2, ..., 2^n-1, that f(z) tends to infinity when z approaches a along the radius of the circle, lim (r>0, r--> 1 from the negative side) f(r*a) = infinity. Conclude that every point of the unit circle is a singularity for f.
Pf.
Hmm, well I first of all don't really even understand the question. We are saying, given an n and given an r that satisfies our limit condition then
f(z) --> infinity? and what exactly is a supposed to be?
(5) Suppose that f(z) is given by a power series and has radius of convergence equal to 1 and there are only poles of first order on the unit circle (no other singularities). Show that the sequence {an} is bounded.
Pf.
? lol
(6) Suppose we wanted to represent the function
f(z) = 1/(1-z^2) + 1/(3-z) by a Laurent series. How many such representations are there? In which region is each of them valid? Find the coefficients aj explicitly for each of these representations.
-
I know that there is more than 1 singularity, but they are isolated singularities. and there are 3 ways to represent a function by a Laurent series, one is by the power series if the function is continuous, however this is not a continuous function, the 2nd is by by terms with negative indice coefficients, and the third is by an infinite series. Need some guidance on this one again.
Thank you!