freckles
  • freckles
\[\text{ Suppose } f \text{ is an analytic function of the complex variable } z=x+i y \text{ given by } \\ f(z)=(2x+3y)+i g(x,y) \\ \text{ where} g(x,y) \text{ is a real-valued function } \\ \text{ of the real variables } x \text{ and } y . \\ \text{ If } g(2,3)=1\text{, then }g(7,3)=? . \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • 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
freckles
  • freckles
There are choices A) -14 B) -9 C) 0 D) 11 E) 18 This question comes from the mathematics gre practice exam.
xapproachesinfinity
  • xapproachesinfinity
hmm interesting, i have some admiration for complex though i have no skills in such things yet lol

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anonymous
  • anonymous
I believe it might have something to do with cancelling the complex part.
xapproachesinfinity
  • xapproachesinfinity
i have a feeling that it is not that hard just kind of a trick
freckles
  • freckles
was kind of thinking that too
freckles
  • freckles
somehow we have to show it is -14
freckles
  • freckles
maybe and I could be totally crazy we could do something with magnitudes of complex numbers
freckles
  • freckles
\[|g(2,3)|=1 \\ |f(2,3)|=\sqrt{170} \\ |g(7,3)|=\sqrt{a^2} \text{ if } g(7,3)=a \\ |f(7,3)|=\sqrt{23^2+g^2(7,3)}\]
freckles
  • freckles
\[f^2(7,3)=23^2+a^2 \]
xapproachesinfinity
  • xapproachesinfinity
hmm
freckles
  • freckles
still playing
xapproachesinfinity
  • xapproachesinfinity
you -14 was the answer! did you just tried it directly?
xapproachesinfinity
  • xapproachesinfinity
said*
freckles
  • freckles
no I looked at the answers
xapproachesinfinity
  • xapproachesinfinity
oh ok
freckles
  • freckles
how did you know -14 was right?
xapproachesinfinity
  • xapproachesinfinity
no i just saw your reply above mentioning that
freckles
  • freckles
I wonder what analytic means
freckles
  • freckles
looking up...
xapproachesinfinity
  • xapproachesinfinity
isn't that related to graphs ?
xapproachesinfinity
  • xapproachesinfinity
don't know why wolfram ignores the imaginary stuff for such functions
freckles
  • freckles
https://www3.nd.edu/~atassi/Teaching/ame60612/Notes/analytic_functions.pdf go to page 2
freckles
  • freckles
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=8&cad=rja&uact=8&ved=0CFMQFjAH&url=https%3A%2F%2Fwww3.nd.edu%2F~atassi%2FTeaching%2Fame60612%2FNotes%2Fanalytic_functions.pdf&ei=LFh3VZrEDcSryASKgoeoBw&usg=AFQjCNHkY1gTLqN3LP3prcv3sWN0hzHCxw&sig2=F-sCq97jHaHPynarc5kPTw
freckles
  • freckles
\[f(x,y)=u(x,y)+i v(x,y) \\ \frac{\partial u}{ \partial x}=\frac{\partial v}{ \partial y} \\ \frac{\partial u}{ \partial y}=-\frac{ \partial v }{\partial x} \\ 2=g_y \\ 3=-g_x\]
anonymous
  • anonymous
Analytic functions refers to something else. One important property of analytic functions are that they are indefinitely differentiable.
xapproachesinfinity
  • xapproachesinfinity
you know it crossed my mind that we need to use calculus lol
freckles
  • 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
xapproachesinfinity
  • xapproachesinfinity
so any analytic function has no differentiability problem is that what you are saying?
freckles
  • freckles
\[2g(x,y)=2y-3x+C(x)-K(y) \\ 2g(2,3)=2(3)-3(2)+C(2)-K(3) \\ g(2,3)=C(2)-K(3)\]
freckles
  • freckles
err...
xapproachesinfinity
  • xapproachesinfinity
oh i see! after checking the theorem in that pdf
freckles
  • freckles
lol still trying to play with this fungus
freckles
  • freckles
and that is equal to 1
freckles
  • freckles
\[1=C(2)-K(3)\]
freckles
  • freckles
\[2g(7,3)=2(3)+3(7)+C(7)-K(3)\]
freckles
  • freckles
\[2g(7,3)=27+C(7)-K(3)\]
freckles
  • freckles
\[2g(7,3)-1=27+C(7)-C(2)\]
anonymous
  • anonymous
Oh yea, use the Cauchy–Riemann equations to solve this.
freckles
  • freckles
\[2g(7,3)=28+C(7)-C(2) \\ g(7,3)=14+\frac{C(7)-C(2)}{2}\]
freckles
  • freckles
oops i made a type-o somewhere above
freckles
  • freckles
\[g(x,y)=2y+C(x) \\ -g(x,y)=3x+K(y) \\ \\ 0=2y+3x+C(x)+K(y)\] \[2g(x,y)=2y-3x+C(x)-K(y) \\ 2g(2,3)=2(3)-3(2)+C(2)-K(3) \\ g(2,3)=C(2)-K(3)\] \[C(2)-K(3)=1 \\ 2g(7,3)=6-21+C(7)-K(3) \\ 2g(7,3)=-15+C(7)-K(3) \\ 2g(7,3)-1=-15+C(7)-K(3)-C(2)+K(3) \\ 2g(7,3)=-14+C(7)-C(2) \\ g(7,3)=-7+\frac{C(7)-C(2)}{2}\]
freckles
  • freckles
I think I'm still making a mistake somewhere
anonymous
  • anonymous
Oh I get it now.
anonymous
  • anonymous
You simply use the Cauchy-Riemann equations above, and calculate the partials. There, you had \[2 = g_{y}\]
anonymous
  • anonymous
The C-R equations are your best bet. It becomes a problem similar to finding potential functions.
anonymous
  • anonymous
and \[3 = -g_{x}\]
anonymous
  • anonymous
Yes, and then you find the potential function by adding the constant term.
anonymous
  • anonymous
So your potential function g(x,y) becomes -3x + 2y + C
anonymous
  • 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.
anonymous
  • anonymous
Therefore, you get \[g(x,y) = -3x+2y+1\]
anonymous
  • anonymous
So just plug in, g(7,3) and you will get -14.
freckles
  • freckles
@Loser66 check this out
freckles
  • 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
freckles
  • freckles
thanks for your help guys

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