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

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

Mathematics
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\[f(2+3i)=4+9+ig(2,3) \\ f(2+3i)=13+i g(2,3) \\ f(2+3i)=13+i \\ f(7+3i)=14+9+ig(7,3) \\ f(7+3i)=23+i g(7,3) \\ \] this as far as I can get so far
There are choices A) -14 B) -9 C) 0 D) 11 E) 18 This question comes from the mathematics gre practice exam.
hmm interesting, i have some admiration for complex though i have no skills in such things yet lol

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Other answers:

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

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