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Non existence of complex functions

Complex Analysis
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Tsakanikas Nickos
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Non existence of complex functions

#1

Post by Tsakanikas Nickos » Thu Jul 14, 2016 7:00 pm

  1. Show that the only entire function \( \displaystyle f \) that satisfies \( \displaystyle \Big| f(z)-i \Big| < \Big| f(z) \Big|e^{\Re(z)} \, , \, \forall z \in \mathbb{C} \) is the constant function \( \displaystyle f(z)=i \, , \, z \in \mathbb{C} \).
  2. Show that a complex polynomial function \( \displaystyle P \) such that \( \displaystyle \Big| {e}^z + P(z) \Big| < e^{\Re(z)} \, , \, \forall z \in \mathbb{C} \) does not exist.
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Re: Non existence of complex functions

#2

Post by Captainjp » Thu Jul 14, 2016 7:01 pm

Tsakanikas Nickos wrote:
  1. Show that the only entire function \( \displaystyle f \) that satisfies \( \displaystyle \Big| f(z)-i \Big| < \Big| f(z) \Big|e^{\Re(z)} \, , \, \forall z \in \mathbb{C} \) is the constant function \( \displaystyle f(z)=i \, , \, z \in \mathbb{C} \).
Since \( f(z)=0 \) is impossible for any \( z\in \mathbb{C} \), we can define the entire function \( g(z)=1-\frac{i}{f(z)} \). By the hypothesis it is \( |g(z)|<e^{\Re(z)} \).
If we restrict the function to the half-plane where \( \Re(z)\geq 0 \), then \( g \) is bounded which by a known fact of complex analysis means it is constant. Obviously just by taking the limit when \( \Re(z)\rightarrow -\infty \) it is \( g=0 \). Also by a known theorem of complex analysis, since \(g \) is constant in a disc and holomorphic, it must be constant in the entire plane. So \(g=0 \) everywhere which implies \( f(z)=i \) everywhere.
Tsakanikas Nickos
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Re: Non existence of complex functions

#3

Post by Tsakanikas Nickos » Thu Jul 14, 2016 7:02 pm

Tsakanikas Nickos wrote: Show that a complex polynomial function \( \displaystyle P \) such that \( \displaystyle \Big| {e}^z + P(z) \Big| < e^{\Re(z)} \, , \, \forall z \in \mathbb{C} \) does not exist.
Suppose that such a polynomial function \( \displaystyle P \) exists. Dividing \( \displaystyle \Big| {e}^z + P(z) \Big| < e^{\Re(z)} \) by \( \displaystyle \big| {e}^{z} \big| \), we have that \( \displaystyle \Big| 1 + {e}^{-z}P(z) \Big| < 1 \, , \, \forall z \in \mathbb{C} \). Therefore, the function \( \displaystyle f(z) = 1 + {e}^{-z}P(z) \) is entire and bounded, so by Liouville's theorem we conclude that \( \displaystyle f \) is constant. Hence, there is a \( \displaystyle c \in \mathbb{C} \) such that \( \displaystyle f(z)=c \, , \, \forall z \in \mathbb{C} \) and thus \( \displaystyle P(z)=(c-1){e}^{z} \, , \, \forall z \in \mathbb{C} \). Observe that \( \displaystyle c \neq 1 \) due to the relationship satisfied by \( \displaystyle P \). Hence, the polynomial function \( \displaystyle P \) is a non zero multiple of the exponential function \( \displaystyle {e}^{z} \). This is a contradiction!
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Re: Non existence of complex functions

#4

Post by S.F.Papadopoulos » Thu Sep 01, 2016 3:05 pm

If we restrict the function to the half-plane where ${\rm Re}(z) \geq 0$ , then $g$ is bounded which by a known fact of complex analysis means it is constant. The above is false. $g(z)=\exp(-z)$.
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