Spring 2020.3, Extras Fall 2009 #complex/qual/completed
- Assume f(z)=∑∞n=0cnzn converges in |z|<R. Show that for r<R,
12π∫2π0|f(reiθ)|2dθ=∞∑n=0|cn|2r2n
- Deduce Liouville’s theorem from (a).
solution:
Computing the LHS: ∫[0,2π]|f(reiθ)|2dθ=∫[0,2π]f(reiθ)¯f(reiθ)dθ=∫[0,2π]∑k≥0ckrkeikθ∑j≥0¯cjrje−ijθdθ=∫[0,2π]∑k,j≥0ck¯cjrk+jei(k−j)θdθ=∑k,j≥0ck¯cjrk+j∫[0,2π]ei(k−j)θdθ=∑k,j≥0ck¯cjrk+jχi=j⋅2π=∑k≥0ck¯ckr2k⋅2π=2π∑k≥0|ck|2r2k, where we’ve used that the series converges uniformly in its radius of convergence to commute sums and integrals.
Now supposing |f(z)|≤M for all z∈C, if f is entire then ∑k≥0ckzk converges for all r, so ∑k≥0|ck|2r2k=12π∫[0,2π]|f(reiθ)|2dθ≤12π∫[0,2π]M2dθ=M2. Thus for all r, |c0|2+|c1|2r2+|c2|2r4+⋯≤M2, and taking r→∞ forces |c1|2=|c2|2=⋯=0. So f(z)=c0 is constant.
FTA via Liouville #complex/exercise/completed
Prove the Fundamental Theorem of Algebra (using complex analysis).
solution:
- Strategy: By contradiction with Liouville’s Theorem
- Suppose p is non-constant and has no roots.
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Claim: 1/p(z) is a bounded holomorphic function on C.
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Holomorphic: clear? Since p has no roots.
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Bounded: for z≠0, write P(z)zn=an+(an−1z+⋯+a0zn).
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The term in parentheses goes to 0 as |z|→∞
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Thus there exists an R>0 such that |z|>R⟹|P(z)zn|≥c:=|an|2.
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So p is bounded below when |z|>R
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Since p is continuous and has no roots in |z|≤R, it is bounded below when |z|≤R.
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Thus p is bounded below on C and thus 1/p is bounded above on C.
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- By Liouville’s theorem, 1/p is constant and thus p is constant, a contradiction.
Entire functions satisfying an inequality #complex/exercise/completed
Find all entire functions that satisfy |f(z)|≥|z|∀z∈C. Prove this list is complete.
- If f is bounded in a neighborhood of a singularity z0, then z0 is removable.
solution:
- Suppose f is entire and define g(z):=zf(z).
- By the inequality, |g(z)|≤1, so g is bounded.
- g potentially has singularities at the zeros Zf:=f−1(0), but since f is entire, g is holomorphic on C∖Zf.
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Claim: Zf={0}.
- If f(z)=0, then |z|≤|f(z)|=0 which forces z=0.
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We can now apply Riemann’s removable singularity theorem:
- Check g is bounded on some open subset D∖{0}, clear since it’s bounded everywhere
- Check g is holomorphic on D∖{0}, clear since the only singularity of g is z=0.
- By Riemann’s removable singularity theorem, the singularity z=0 is removable and g has an extension to an entire function ˜g.
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By continuity, we have |˜g(z)|≤1 on all of C
- If not, then |˜g(0)|=1+ε>1, but then there would be a domain Ω⊆C∖{0} such that 1<|˜g(z)|≤1+ε on Ω, a contradiction.
- By Liouville, ˜g is constant, so ˜g(z)=c0 with |c0|≤1
- Thus f(z)=c−10z:=cz where |c|≥1
Thus all such functions are of the form f(z)=cz for some c∈C with |c|≥1.
Entire functions with an asymptotic bound #complex/exercise/completed
Find all entire functions satisfying |f(z)|≤|z|12 for |z|>10.
solution:
Since f is entire, take a Laurent expansion at z=0, so f(z)=∑k≥0ckzk where 2πik!ck=f(k)(0) by Cauchy’s integral formula. Take a Cauchy estimate on a disc of radius R>10: |ck|≤k!2π∫|z|=R|f(ξ)(ξ−0)k+1|dξ≤k!2π∫|z|=R|ξ|12|ξ|k+1dξ=k!2π⋅1Rk+12⋅2πR=O(1/Rk−12). So in particular, if k≥1 then k−12>0 and ck=0. This forces f=c0 to be constant.
Tie’s Extra Questions: Fall 2009 #complex/exercise/completed
Let f(z) be entire and assume values of f(z) lie outside a bounded open set Ω. Show without using Picard’s theorems that f(z) is a constant.
solution:
We have |f(z)|≥M for some M, so |1/f(z)|≤M−1 is bounded, and we claim it is entire as well. This follows from the fact that 1/f has singularities at the zeros of f, but these are removable since 1/f is bounded in every neighborhood of each such zero. So 1/f extends to a holomorphic function. But now 1/f=c is constant by Liouville, which forces f=1/c to be constant.
Tie’s Extra Questions: Fall 2015 #complex/exercise/completed
Let f(z) be bounded and analytic in C. Let a≠b be any fixed complex numbers. Show that the following limit exists: lim
Use this to show that f(z) must be a constant (Liouville’s theorem).
solution:
Apply PFD and use that f is holomorphic to apply Cauchy’s formula over a curve of radius R enclosing a and b: \begin{align*} \int_\gamma {f(z) \over (z-a)(z-b)}\,dz &= \int_\gamma f(z)\qty{{a-b \over z-a} + {b-a\over z-b} } \,dz\\ &= (a-b)^{-1}\int_\gamma {f(z) \over z-a} \,dz+ (b-a)^{-1}\int_\gamma {f(z) \over z-b}\,dz\\ &= (a-b)^{-1}\cdot 2\pi i f(a) + (b-a)\cdot 2\pi i f(b)\\ &= 2\pi i\qty{f(a) - f(b) \over a-b } .\end{align*} Since f is bounded, this number is finite and independent of R, so taking R\to\infty preserves this equality. On the other hand, if {\left\lvert {f(z)} \right\rvert}\leq M, then we can estimate this integral directly as \begin{align*} I \leq \int_{{\left\lvert {z} \right\rvert} = R} {M \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert} } = {M\cdot 2\pi R \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert}} \ll {1\over R} \to 0 ,\end{align*} which forces f(a) =f(b). Since a, b were arbitrary, f must be constant.