Loser66
  • Loser66
Confuse!! Please, help Problem: Where is the function differentiable? \(x^2+y^2 +2ixy\) Where is it analytic?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Loser66
  • Loser66
\(U_x = 2x = V_y \\U_y = 2y = -V_x = -2y \iff y =0\) 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
  • anonymous
hmm you sure this is right?
Loser66
  • Loser66
What is your suspicion?

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anonymous
  • anonymous
checking the partials although it can't be that hard, i think it must be of by a minus sign somewhere
Loser66
  • Loser66
\(U(x,y) = x^2 +y^2 \\V(x,y) = 2xy\) \(U_x = 2x\\U_y= 2y\\V_x = 2y\\V_y = 2x\) I am sure about that.
anonymous
  • anonymous
then you are off by a minus sign for cauchy riemman right?
Loser66
  • Loser66
\(U_x = V_y=2x\\U_y = -V_x\iff 2y = -2y \iff y =0\) I am sure about that also.
Loser66
  • Loser66
Yes,
anonymous
  • anonymous
for cauchy riemann \(U_y=-V_x\)
Loser66
  • Loser66
Yes
anonymous
  • anonymous
but is isn't
Loser66
  • Loser66
Hence, the conclusion is the function is differentiable when y =0
Loser66
  • Loser66
That is it is differentiable on Real axis, right?
anonymous
  • anonymous
yeah on the real axis it is \(f(x)=x^2\)
Loser66
  • Loser66
But 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"
Loser66
  • Loser66
It is not f (z)
anonymous
  • anonymous
differentiable means the derivative exists from any direction
Loser66
  • Loser66
f(z) = x^2 +y^2 +2ixy,
Loser66
  • Loser66
and the whole thing differentiable at the point on Real axis only. We can't take off y from f, right?
anonymous
  • anonymous
no 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
  • anonymous
btw 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 \(z\)
Loser66
  • Loser66
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anonymous
  • anonymous
which is probably not not the case here, \[f(z)=z^2=x^2-y^2+2xyi\]
anonymous
  • anonymous
i am pretty sure the answer is no the limit (difference quotient) has to exist from any direction
anonymous
  • anonymous
Another 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
  • Loser66
I got it. Hence, from neighborhood of a point on Real axis, f is not differentiable,
anonymous
  • anonymous
the Cauchy-Riemann 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
  • anonymous
this is the connection to 'being written as a function of only \(z\), not \(\bar z\)' @satellite73

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