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f(z) = { x.y^2.(x + iy) + (x^2 + y^2), z != 0
{ 0 , z = 0

z = x + iy ; i = 'iota'

to show that f(z) is not analytic at z = 0, although C-R equations are satisfied at the origin.

Plzzz help!!

2006-12-20 05:50:15 · 2 answers · asked by Sunny 3 in Science & Mathematics Mathematics

srry!
the f(z) at z != 0 is = x.y^2.(x + iy) + (x^2 + y^4)

it's not y^2, it's y^4.

2006-12-20 05:52:28 · update #1

I've already stated that CR test is satisfied

2006-12-20 06:10:52 · update #2

2 answers

Perform a Cauchy-Riemann test.

2006-12-20 06:08:32 · answer #1 · answered by Edward 7 · 0 1

that's somewhat somewhat comparable to differentiability in distinctive genuine dimensions. The C-R equations take care of partials, so we can't get the great image from them. although, if we additionally be attentive to that the function is non-end, then all of us be attentive to that this is analytic. in short: Analytic <-> (C-R holds and non-end) See the source occasion 3.6 for a function the place C-R holds at a element, however the function isn't analytic in any open community of the element. @JB: It looks such as you have got referenced wikipedia. the priority is that the wikipedia article demands u and v to be differentiable => u and v non-end => f non-end. i'm thinking that the wiki website desires somewhat cleanup to stay away from this confusion... 2d Edit: regrettably i've got not got my diagnosis texts with me on the 2nd. i became utilising "no longer analytic at a element" to point "no longer analytic in any open community of the element", this is in basic terms the assumption of a singularity. I replaced it with the intention to stay away from to any extent further confusion.

2016-12-11 12:57:12 · answer #2 · answered by Anonymous · 0 0

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