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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.