That f(z) is constant if and only if g(z) is constant, and because g(z) is bounded and entire, by Liouvilles theorem it is constant, which implies f(z) is constant
Ahhh, no the denominator can't vanish because we are dividing 1000 by an inequality which is greater or equal to 1000, therefore, the maximum value g(z) can take is 1 if |1000i+f(z)|=1000.
So, as the denominator does not vanish it must be analytic everywhere
I would say g(z) is differentiable on the whole C - {-1000i}, and so would be entire within the region {z:Imz > - 1000}. And is bounded above at g(z)=1,Therefore, by Liouvilles Theorem f(z) is a constant? i.e. if f(z)=0 then g(z)=i < 1…?
Homework Statement
Show that if f is an entire function that satisfies
|1000i + f(z)| ≥ 1000, for all z ∈ C, then f is constant.
Homework Equations
(Hint: Consider the function g(z) = 1000/1000i+f(z) , and apply Liouville’s Theorem.)
The Attempt at a Solution
Ok, so I assume that as f is...
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I'm looking for some light reading for this summer after my exams, and came across this book called "Zombies and Calculus"... looks brilliant. Just want to know if anyone has read it, or any other good math books out there for light reading.
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Hi all from Sunny Reading in Berkshire UK, I've just started my final year undergraduate Maths, but will be taking 2 years over it as I'm a part timer, at the moment the 2 courses I'm doing are complex analysis, and fluid mechanics. Hopefully if I pass them, I will be starting QM and Probability...