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

- Loser66

- schrodinger

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- 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

hmm you sure this is right?

- Loser66

What is your suspicion?

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## More answers

- anonymous

checking the partials although it can't be that hard, i think it must be of by a minus sign somewhere

- 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

then you are off by a minus sign for cauchy riemman right?

- Loser66

\(U_x = V_y=2x\\U_y = -V_x\iff 2y = -2y \iff y =0\)
I am sure about that also.

- Loser66

Yes,

- anonymous

for cauchy riemann \(U_y=-V_x\)

- Loser66

Yes

- anonymous

but is isn't

- Loser66

Hence, the conclusion is the function is differentiable when y =0

- Loser66

That is it is differentiable on Real axis, right?

- anonymous

yeah on the real axis it is \(f(x)=x^2\)

- 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

It is not f (z)

- anonymous

differentiable means the derivative exists from any direction

- Loser66

f(z) = x^2 +y^2 +2ixy,

- Loser66

and the whole thing differentiable at the point on Real axis only. We can't take off y from f, right?

- 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

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

|dw:1443751466835:dw|

- anonymous

which is probably not not the case here, \[f(z)=z^2=x^2-y^2+2xyi\]

- anonymous

i am pretty sure the answer is no
the limit (difference quotient) has to exist from any direction

- 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

I got it. Hence, from neighborhood of a point on Real axis, f is not differentiable,

- 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

this is the connection to 'being written as a function of only \(z\), not \(\bar z\)' @satellite73

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