Bounds for Modulus of e^z on Circle x^2 + y^2 - 2x - 2y - 2 = 0
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Theorem
Consider the circle $C$ embedded in the complex plane defined by the equation:
- $x^2 + y^2 - 2 x - 2 y - 2 = 0$
Let $z = x + i y \in \C$ be a point lying on $C$.
Then:
- $e^{-1} \le \cmod {e^z} \le e^3$
Proof
\(\ds x^2 + y^2 - 2 x - 2 y - 2\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {\paren {x - 1}^2 - 1} + \paren {\paren {y - 1}^2 - 1} - 2\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x - 1}^2 + \paren {y - 1}^2\) | \(=\) | \(\ds 4\) |
This defines a circle whose center is at $1 + i$ and whose radius is $2$.
From Modulus of Exponential is Exponential of Real Part:
- $\cmod {e^z} = e^x$
If $z \in C$ then from the geometry of the circle $C$:
- $-1 \le x \le 3$
Then from Exponential is Strictly Increasing:
- $e^{-1} \le e^x \le e^3$
Hence the result.
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 4$. Elementary Functions of a Complex Variable: Exercise $5$