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

\(\displaystyle x^2 + y^2 - 2 x - 2 y - 2\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {\paren {x - 1}^2 - 1} + \paren {\paren {y - 1}^2 - 1} - 2\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x - 1}^2 + \paren {y - 1}^2\) \(=\) \(\displaystyle 4\)

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