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Holomorphic Function

  1. Mar 20, 2009 #1
    Is f(z) = [itex](\bar{z})^2[/itex] holomorphic in [itex]\mathbb{C}[/itex]? Is it differentiable in the complex sense at any point [itex]z_0 \in \mathbb{C}[/itex]? If it is, find [itex]f’(z_0)[/itex].

    for the first part i said it was not holomorphic on the whole complex plane as [itex]f(z)=(x-iy)^2=x^2-2ixy-y^2=u(x,y)+iv(x,y)[/itex]

    and so [itex]\frac{\partial{u}}{\partial{x}}=2x,\frac{\partial{u}}{\partial{y}}=-2y,\frac{\partial{v}}{\partial{x}}=-2y,\frac{\partial{v}}{\partial{y}}=-2x[/itex] meaning the Cauchy Riemann equations are not satisfied unless [itex](x,y)=(0,0)[/itex] i.e. it is holomorphic at the origin of the copmlex plane and therefore it may be differentiable at this point - this leads us to the next part of the question.

    so for f to be differentiable at a particular [itex]z_0 \in \mathbb{C}[/itex] it must be holomorphic at that point, let us investigate what happens at [itex]z_0=(0,0)[/itex]

    [itex]\mathop {\lim }\limits_{h \to 0} \frac{f(z_0+h)-f(z_0)}{h} = \mathop {\lim }\limits_{h \to 0} \frac{f(h)}{h}=\mathop {\lim }\limits_{h \to 0} \frac{\bar{h}^2}{h} \rightarrow 0[/itex].
    now as this limit exists, f must be differentiable at 0

    the derivative is given by [itex]f'(z)=\frac{\partial{u}}{\partial{x}}+i \frac{\partial{v}}{\partial{x}}=0[/itex]

    i'm a little uncertain of the final answer so would like someone to check over my working if possible. cheers.
  2. jcsd
  3. Mar 21, 2009 #2
    You are right, the complex differential exists only at the origin and it is 0. By the way, since you alread computed the partial derivatives and found the Cauchy-Riemann equations to be satisfied at 0, this already implies that the complex derivative (i.e. the limit of the difference quotient) exists at 0.

    However, the definition of "holomorphic" requires an open set U of the complex plane and the complex derivative to exists for all points in U. Otherwise the familiar properties, like existence and convergence of the taylor series, would fail. So it does not make sense to say that f is holomorphic at the origin, since a single point is not open.
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