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Loser66
 one year ago
Confuse!! Please, help
Problem: Where is the function differentiable? \(x^2+y^2 +2ixy\)
Where is it analytic?
Loser66
 one year ago
Confuse!! Please, help Problem: Where is the function differentiable? \(x^2+y^2 +2ixy\) Where is it analytic?

This Question is Closed

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(U_x = 2x = V_y \\U_y = 2y = V_x = 2y \iff y =0\) Hence the function is differentiable on x axis. My question: is it not that it is analytic on x axis also? why the answer is : "No where" ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm you sure this is right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0What is your suspicion?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0checking the partials although it can't be that hard, i think it must be of by a minus sign somewhere

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(U(x,y) = x^2 +y^2 \\V(x,y) = 2xy\) \(U_x = 2x\\U_y= 2y\\V_x = 2y\\V_y = 2x\) I am sure about that.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then you are off by a minus sign for cauchy riemman right?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0\(U_x = V_y=2x\\U_y = V_x\iff 2y = 2y \iff y =0\) I am sure about that also.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for cauchy riemann \(U_y=V_x\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Hence, the conclusion is the function is differentiable when y =0

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0That is it is differentiable on Real axis, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah on the real axis it is \(f(x)=x^2\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0But for part b) Where is it analytic, From above it is differentiable on Real, it should be analytic on Real also, but the answer is "No where"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0differentiable means the derivative exists from any direction

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0f(z) = x^2 +y^2 +2ixy,

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0and the whole thing differentiable at the point on Real axis only. We can't take off y from f, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no just like on the real line differentiable means the limit exists from the left and the right, in complex plane the limit must exist from any direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw if they didn't mention it, as i remember correctly "analytic" means differentiable, but also means you can write the function as a function of \(z\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which is probably not not the case here, \[f(z)=z^2=x^2y^2+2xyi\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i am pretty sure the answer is no the limit (difference quotient) has to exist from any direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another way of repeating what @satellite73 has already stated is that a function can't be analytic on a line  it can only be analytic on a disk. It can be differentiable on a line, but to be analytic the function has to be differentiable no matter how you approach the point in the complex plane.

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0I got it. Hence, from neighborhood of a point on Real axis, f is not differentiable,

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the CauchyRiemann equations that govern complex differentiability of \(f(x+iy)=u(x,y)+iv(x,y)\) basically require that we have the following condition on a Wirtinger derivative: $$\frac{\partial f}{\partial\bar z}=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is the connection to 'being written as a function of only \(z\), not \(\bar z\)' @satellite73
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