find the analytic function f(z)=u(x,y)+iv(x,y) such that u+v=(x-y)(x^2+4xy+y^2)
split the fraction maybe? or decomp it?
what do you mean by split the fraction ??
@amistre64 what do you mean by split the fraction ???
just a thought (a+b)/k = a/k + b/k
ok that's partial fraction how am I supposed to use it here ?? am I missing something ?
partial would have use factor k, i havent seen a way to factor k so i suggested the splits
Hmm I don't remember complex very well... Oldrin could probably explain this well. If f is analytic, then it must satisfy the Cauchy Reimann equations, yes? Do we want to make use of those here maybe? \(\large\rm u_x=v_y\) \(\large\rm u_y=-v_x\)
is u+v division or multiplication? these old eyes get tired
on the right side? multiplication :)
if multiplication, then the sum of 2 things is prolly better expressed as the sum of other things and so distributing it all out might be beneficial ... but i got no cauchy reimann tricks to play with :)
u+v=(x-y)(x^2+4xy+y^2) (u+v)/(x-y) = x^2+4xy+y^2 x^2 +4xy +y^2 ------------------ x-y| [x^3] [+3x^2y] (x^3 -x^2y) ------------- 4x^2y [-3xy^2] (4x^2y -4xy^2) --------------- xy^2 [-y^3] (xy^2-y^3) u+v = x^3 +3x^2y -3xy^2 -y^3
yes the cauchy reimann equation usually problems go like this it gives me u or v and says that it satisfies cauchy reimann equation then using differentiation I get Ux then by integration I get V but never gave me the sum of U+V before
i just worked it 'backwards' to determine what u+v needs to be in order to be divided by (x-y) and gives us (x^2 + 4xy +y^2)
can we split that apart into an appropriate u and v that meet the couchy?
lol, i spose distribution would have sufficed, but when i got an idea in my head its hard to let it go
I found this by searching the internet but can't make sense out of it but it's supposed to help me
yeah, that wouldnt have helped me either :/
lol just asked my colleagues on our facebook group one of them hit me with a killer comment
lol
heheh, they are still on page 14 arent they :)
I think they don't even know that there is a textbook at all
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