How do you show that the function w=zez― is nowhere analytic?

Given W=z.ez′ where z’ denotes complex conjugate of the variable,z=x+iy. Now, W=z.ez′⇒ln⁡(W)=ln⁡(z)+z′ which further implies that(1w)(dWdz)=(1z)+dz′/dz. Obviously dWdz does not exist at any point z of the complex plane as (dz′dz) does not exist anywhere.

Expert Answer to "How do you show that the function w=zez―..."

John Stewart

Answered question

2022-01-17

How do you show that the function

w = z e^{\overset{―}{z}}

is nowhere analytic?

Answer & Explanation

Neunassauk8

Beginner2022-01-18Added 30 answers

w = (x + i y) e^{x - i y} = (x + i y) e^{x} (\cos (y) - i \sin (y))

w = u (x, y) + i v (x, y)

where

u (x, y) = e^{x} (x \cos (y) + y \sin (y))

v (x, y) = e^{x} (y \cos (y) - x \sin (y))

Then

u_{x} = e^{x} \cos (y) + e^{x} (x \cos (y) + y \sin (y))

and

v_{y} = e^{x} (- y \sin (y) + \cos (y) - x \cos (y))

.
Since

u_{x} = v_{y}

only at (0,0), the Cauchy-Riemann Equations are not satisfied on any open set implying that this function is nowhere analytic.
Note. One may also observe that this function is not analytic because

\frac{\partial w}{\partial \overset{―}{z}} = 0

only at

z = 0

puhnut1m

Beginner2022-01-19Added 33 answers

Given $W = z . e^{z^{'}}$ where z’ denotes complex conjugate of the variable,
$z = x + i y$ . Now, $W = z . e^{z^{'}} \Rightarrow \ln (W) = \ln (z) + z^{'}$ which further implies that

$(\frac{1}{w}) (\frac{d W}{d z}) = (\frac{1}{z}) + d z^{'} / d z$ . Obviously $\frac{d W}{d z}$ does not exist at any point z of the complex plane as ( $\frac{d z^{'}}{d z}$ ) does not exist anywhere.

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