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Math Help - Complex Analysis: Show exp|z| not analytic

  1. #1
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    Complex Analysis: Show exp|z| not analytic

    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.
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  2. #2
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    Quote Originally Posted by pineapple89 View Post
    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 z = u(x, y) + i\,v(x, y)

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

    This is a REAL function, not complex.

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

    So we can write it as

    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 V will be 0, while the partial derivatives of U won't always be.

    Therefore the Cauchy-Riemann equations will not be satisfied, and the function is not analytic.
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  3. #3
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    Quote Originally Posted by pineapple89 View Post
    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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