## freckles one year ago $\text{ Suppose } f \text{ is an analytic function of the complex variable } z=x+i y \text{ given by } \\ f(z)=(2x+3y)+i g(x,y) \\ \text{ where} g(x,y) \text{ is a real-valued function } \\ \text{ of the real variables } x \text{ and } y . \\ \text{ If } g(2,3)=1\text{, then }g(7,3)=? .$

1. freckles

$f(2+3i)=4+9+ig(2,3) \\ f(2+3i)=13+i g(2,3) \\ f(2+3i)=13+i \\ f(7+3i)=14+9+ig(7,3) \\ f(7+3i)=23+i g(7,3) \\$ this as far as I can get so far

2. freckles

There are choices A) -14 B) -9 C) 0 D) 11 E) 18 This question comes from the mathematics gre practice exam.

3. xapproachesinfinity

hmm interesting, i have some admiration for complex though i have no skills in such things yet lol

4. anonymous

I believe it might have something to do with cancelling the complex part.

5. xapproachesinfinity

i have a feeling that it is not that hard just kind of a trick

6. freckles

was kind of thinking that too

7. freckles

somehow we have to show it is -14

8. freckles

maybe and I could be totally crazy we could do something with magnitudes of complex numbers

9. freckles

$|g(2,3)|=1 \\ |f(2,3)|=\sqrt{170} \\ |g(7,3)|=\sqrt{a^2} \text{ if } g(7,3)=a \\ |f(7,3)|=\sqrt{23^2+g^2(7,3)}$

10. freckles

$f^2(7,3)=23^2+a^2$

11. xapproachesinfinity

hmm

12. freckles

still playing

13. xapproachesinfinity

you -14 was the answer! did you just tried it directly?

14. xapproachesinfinity

said*

15. freckles

no I looked at the answers

16. xapproachesinfinity

oh ok

17. freckles

how did you know -14 was right?

18. xapproachesinfinity

no i just saw your reply above mentioning that

19. freckles

I wonder what analytic means

20. freckles

looking up...

21. xapproachesinfinity

isn't that related to graphs ?

22. xapproachesinfinity

don't know why wolfram ignores the imaginary stuff for such functions

23. freckles
24. freckles
25. freckles

$f(x,y)=u(x,y)+i v(x,y) \\ \frac{\partial u}{ \partial x}=\frac{\partial v}{ \partial y} \\ \frac{\partial u}{ \partial y}=-\frac{ \partial v }{\partial x} \\ 2=g_y \\ 3=-g_x$

26. anonymous

Analytic functions refers to something else. One important property of analytic functions are that they are indefinitely differentiable.

27. xapproachesinfinity

you know it crossed my mind that we need to use calculus lol

28. freckles

$g(x,y)=2y+C(x) \\ -g(x,y)=3x+K(y) \\ \\ 0=2y+3x+C(x)+K(y)$ where C(x) is a function of x and K(y) is a function of y

29. xapproachesinfinity

so any analytic function has no differentiability problem is that what you are saying?

30. freckles

$2g(x,y)=2y-3x+C(x)-K(y) \\ 2g(2,3)=2(3)-3(2)+C(2)-K(3) \\ g(2,3)=C(2)-K(3)$

31. freckles

err...

32. xapproachesinfinity

oh i see! after checking the theorem in that pdf

33. freckles

lol still trying to play with this fungus

34. freckles

and that is equal to 1

35. freckles

$1=C(2)-K(3)$

36. freckles

$2g(7,3)=2(3)+3(7)+C(7)-K(3)$

37. freckles

$2g(7,3)=27+C(7)-K(3)$

38. freckles

$2g(7,3)-1=27+C(7)-C(2)$

39. anonymous

Oh yea, use the Cauchy–Riemann equations to solve this.

40. freckles

$2g(7,3)=28+C(7)-C(2) \\ g(7,3)=14+\frac{C(7)-C(2)}{2}$

41. freckles

oops i made a type-o somewhere above

42. freckles

$g(x,y)=2y+C(x) \\ -g(x,y)=3x+K(y) \\ \\ 0=2y+3x+C(x)+K(y)$ $2g(x,y)=2y-3x+C(x)-K(y) \\ 2g(2,3)=2(3)-3(2)+C(2)-K(3) \\ g(2,3)=C(2)-K(3)$ $C(2)-K(3)=1 \\ 2g(7,3)=6-21+C(7)-K(3) \\ 2g(7,3)=-15+C(7)-K(3) \\ 2g(7,3)-1=-15+C(7)-K(3)-C(2)+K(3) \\ 2g(7,3)=-14+C(7)-C(2) \\ g(7,3)=-7+\frac{C(7)-C(2)}{2}$

43. freckles

I think I'm still making a mistake somewhere

44. anonymous

Oh I get it now.

45. anonymous

You simply use the Cauchy-Riemann equations above, and calculate the partials. There, you had $2 = g_{y}$

46. anonymous

The C-R equations are your best bet. It becomes a problem similar to finding potential functions.

47. anonymous

and $3 = -g_{x}$

48. anonymous

Yes, and then you find the potential function by adding the constant term.

49. anonymous

So your potential function g(x,y) becomes -3x + 2y + C

50. anonymous

But you are given an initial condition, g(2,3)=1. Plugging it back into our potential function, you will find that C = 1.

51. anonymous

Therefore, you get $g(x,y) = -3x+2y+1$

52. anonymous

So just plug in, g(7,3) and you will get -14.

53. freckles

@Loser66 check this out

54. freckles

I will still need to look at that theorem thing because I think I was using a little wrong. lol like I thought I was supose to write g_x=-3 implies g(x,y)=-3x+K(y) where K is a function of y and g_y=2 implies g(x,y)=2y+C(x) where C is a function of x but I kept getting ugly stuff

55. freckles

thanks for your help guys