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

- freckles

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- 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

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

- xapproachesinfinity

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

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## More answers

- anonymous

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

- xapproachesinfinity

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

- freckles

was kind of thinking that too

- freckles

somehow we have to show it is -14

- freckles

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

- 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

\[f^2(7,3)=23^2+a^2 \]

- xapproachesinfinity

hmm

- freckles

still playing

- xapproachesinfinity

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

- xapproachesinfinity

said*

- freckles

no I looked at the answers

- xapproachesinfinity

oh ok

- freckles

how did you know -14 was right?

- xapproachesinfinity

no i just saw your reply above mentioning that

- freckles

I wonder what analytic means

- freckles

looking up...

- xapproachesinfinity

isn't that related to graphs ?

- xapproachesinfinity

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

- freckles

https://www3.nd.edu/~atassi/Teaching/ame60612/Notes/analytic_functions.pdf
go to page 2

- 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

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

- anonymous

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

- xapproachesinfinity

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

- 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

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

- 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

err...

- xapproachesinfinity

oh i see! after checking the theorem in that pdf

- freckles

lol still trying to play with this fungus

- freckles

and that is equal to 1

- freckles

\[1=C(2)-K(3)\]

- freckles

\[2g(7,3)=2(3)+3(7)+C(7)-K(3)\]

- freckles

\[2g(7,3)=27+C(7)-K(3)\]

- freckles

\[2g(7,3)-1=27+C(7)-C(2)\]

- anonymous

Oh yea, use the Cauchyâ€“Riemann equations to solve this.

- freckles

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

- freckles

oops i made a type-o somewhere above

- 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

I think I'm still making a mistake somewhere

- anonymous

Oh I get it now.

- anonymous

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

- anonymous

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

- anonymous

and \[3 = -g_{x}\]

- anonymous

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

- anonymous

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

- 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

Therefore, you get \[g(x,y) = -3x+2y+1\]

- anonymous

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

- freckles

@Loser66
check this out

- 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

thanks for your help guys

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