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

1. 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" ?

2. anonymous

hmm you sure this is right?

3. Loser66

4. anonymous

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

5. 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.

6. anonymous

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

7. Loser66

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

8. Loser66

Yes,

9. anonymous

for cauchy riemann $$U_y=-V_x$$

10. Loser66

Yes

11. anonymous

but is isn't

12. Loser66

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

13. Loser66

That is it is differentiable on Real axis, right?

14. anonymous

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

15. 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"

16. Loser66

It is not f (z)

17. anonymous

differentiable means the derivative exists from any direction

18. Loser66

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

19. Loser66

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

20. 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

21. 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 $$z$$

22. Loser66

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23. anonymous

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

24. anonymous

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

25. 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.

26. Loser66

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

27. 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: $$\frac{\partial f}{\partial\bar z}=0$$

28. anonymous

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