Complex Analysis: Show exp|z| not analytic

Mar 2010
13
0
Hi there,

With z = u(x,y) + iv(x,y) I have to show that exp|z| is not analytic. so I have

exp|z| = exp{(u^2 + v^2)^1/2}

as |z| is sqrt(u^2 + v^2).


I know that showing the Cauchy Riemann functions aren't satisfied will prove it's not analytic, but I'm not sure how to introduce this.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Hi there,

With z = u(x,y) + iv(x,y) I have to show that exp|z| is not analytic. so I have

exp|z| = exp{(u^2 + v^2)^1/2}

as |z| is sqrt(u^2 + v^2).


I know that showing the Cauchy Riemann functions aren't satisfied will prove it's not analytic, but I'm not sure how to introduce this.
If \(\displaystyle z = u(x, y) + i\,v(x, y)\)

then \(\displaystyle |z| = \sqrt{[u(x, y)]^2 + [v(x, y)]^2}\).

This is a REAL function, not complex.

Therefore \(\displaystyle e^{|z|} = e^{\sqrt{[u(x, y)]^2 + [v(x, y)]^2}}\) is also real.

So we can write it as

\(\displaystyle Z = e^{\sqrt{[u(x, y)]^2 + [v(x, y)]^2}} + 0i = U + i\,V\).


You should be able to see that all the partial derivatives of \(\displaystyle V\) will be \(\displaystyle 0\), while the partial derivatives of \(\displaystyle U\) won't always be.

Therefore the Cauchy-Riemann equations will not be satisfied, and the function is not analytic.
 
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Opalg

MHF Hall of Honor
Aug 2007
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2,789
Leeds, UK
Hi there,

With z = u(x,y) + iv(x,y) I have to show that exp|z| is not analytic. so I have

exp|z| = exp{(u^2 + v^2)^1/2}

as |z| is sqrt(u^2 + v^2).


I know that showing the Cauchy Riemann functions aren't satisfied will prove it's not analytic, but I'm not sure how to introduce this.
You have set this up wrongly. Instead of the equation in red, you should start with the equations z = x + iy and exp|z| = u(x,y) + iv(x,y). But then notice that exp|z| is always real, so v(x,y) = 0. Now use the Cauchy–Riemann equations as in Prove_It's comment to show that exp|z| cannot be analytic.
 
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