Show where the functions is anlaytic and differentiable

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SUMMARY

The function f(z) = x³ + i(1 - y)³ is shown to be neither analytic nor differentiable in any neighborhood of the complex plane. The Cauchy-Riemann equations, which are necessary for a function to be analytic, are evaluated and found to hold only at the point (0,0). Specifically, the equations ux = vy and uy = -vx yield results that indicate the function fails to meet the criteria for analyticity elsewhere.

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Homework Statement



z→x3+ i(1 - y)3: Show where the functions is analytic and differentiable.

Homework Equations





The Attempt at a Solution


For a function to be analytic cauchy-riemann equations must hold.. so
ux = vy and uy = -vx

Now f(z) = x3 + i(1 - y)3 is already in the form u(x,y) + iv(x,y) with u(x,y) = x3 and v(x,y) = (1 - y)3 So:

ux = 3x2 ; vy = -3(1 - y)2 and
uy = 0 ; -vx = 0

Does this then mean cauchy-riemann does not hold anywhere but 0 so f is not differentiable on a neighbourhood of any point in C which means it is nowhere analytic.
 
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From what you wrote down it doesn't look like the Cauchy Riemann equations hold at x=y=0
 

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