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Prove that [itex]f(z) = |z|[/itex] is not holomorphic.

  1. Oct 21, 2007 #1
    Prove that [itex]f(z) = |z|[/itex] is not holomorphic.

    We have [itex]f(z) = \sqrt {x^2 + y^2} + i0[/itex], so [itex]\frac {\partial{u}}{\partial{x}} = x(x^2 + y^2)^{ - \frac {1}{2}},\frac {\partial{u}}{\partial{y}} = y(x^2 + y^2)^{ - \frac {1}{2}}, \frac {\partial{v}}{\partial{x}} = \frac {\partial{v}}{\partial{y}} = 0[/itex]

    One sees that if either x or y, but not both, were zero, either condition would be satisfied and the other not. So there are no such points, for which the Riemann conditions would hold.

    Since holomorphic functions are complex-differentiable, and f is not differentiable, it follows that f is not holomorphic.

    It strikes me that the equivalent condition [itex]\frac {\partial{f}}{\partial{\bar{z}}} = 0[/itex] is not trustworthy, because using only this one could say that f is differentiable only in the origin, which is false if we work with Riemann. Why is it so?
  2. jcsd
  3. Oct 22, 2007 #2

    matt grime

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    But your other definition is satisfied at the origin too. I'm not sure I see your point.
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