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## Non existence of complex functions

Complex Analysis
Tsakanikas Nickos
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### Non existence of complex functions

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.
Captainjp
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### Re: Non existence of complex functions

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

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!
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. $g(z)=\exp(-z)$.