Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

was kind of thinking that too

Answer 7

somehow we have to show it is -14

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

hmm

Answer 12

still playing

Answer 13

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

Answer 14

said*

Answer 15

no I looked at the answers

Answer 16

oh ok

Answer 17

how did you know -14 was right?

Answer 18

no i just saw your reply above mentioning that

Answer 19

I wonder what analytic means

Answer 20

looking up...

Answer 21

isn't that related to graphs ?

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

err...

Answer 32

oh i see! after checking the theorem in that pdf

Answer 33

lol still trying to play with this fungus

Answer 34

and that is equal to 1

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

oops i made a type-o somewhere above

Answer 42

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

Answer 43

I think I'm still making a mistake somewhere

Answer 44

Oh I get it now.

Answer 45

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

Answer 46

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

Answer 47

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

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

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

Answer 53

@Loser66
check this out

Answer 54

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

Answer 55

thanks for your help guys