Proving Analytic Functions with Complex Variables

[squaremeplz]

Homework Statement

Suppose f(z) is an analytic function on domain D, and suppose that, for all z in D, we have 2*Re(f(z)) + 3*Im(f(z))=12. prove that f(z) must be a constant.

Homework Equations

The Attempt at a Solution

ok, I am drawing somewhat of blank with this one but I am guessing it has something to do with the partial derivatives.

since f(x +yi) = u(x,y) + i*v(x,y)

i rewrite the equations as 2*u(x,y) * 3*v(x,y) = 12

since f(z) is analytics on D, i know that

u_x' = v_y' and u_y' = - v_x'

but if I differentiate both sides of 2*(u,x) * 3*v(x,y) = 12 with respect to y and x I get a slope of 0 in each case, i.e

2*u_x' + 3*v_x' = 0

and

2*u_y' + 3*v_y' = 0

the only solution for these two equations to hold is one where f(z) is constant.

Is this correct?
any help is appreciated.

 

[e(ho0n3]

2*u_x' + 3*v_x' = 0

and

2*u_y' + 3*v_y' = 0

the only solution for these two equations to hold is one where f(z) is constant.

Can you elaborate on this?

 

[squaremeplz]

hmm elaborate how? I'm guessing I must've gotten something right.

when f(z) is constant f(z) = a + bi

since

f(x + iy) = a + bi for all x,y in R

and since

1) 2*u_x' + 3*v_x' = 0

and

2) 2*u_y' + 3*v_y' = 0

since it's analytic we know that

u_x' = v_y' and u_y' = - v_x'

but now, the only solutions for 1 and 2 to hold MUST be 0 ad f(z) is constant.

is this more clear, more importantly.. is it right? thanks!

 

[xaos]

you need to expand on this. all you've said is "bunch of equations, and so f(z)=const". to show that f(z)=const you need to show that f(z) is independent of z. remember, these four equations hold for all z in D.

 

[squaremeplz]

1) 2*u_x' + 3*v_x' = 0

and

2) 2*u_y' + 3*v_y' = 0

since it's analytic we know that

3) u_x' = v_y' and u_y' = - v_x'

therefore, the partial derivatives need to satisfy 1, 2,3 and by substituting the derivatives from 3 into 1 and 2 , we see the following eq's also need to hold:

4) is also 2*v_y' - 3*u_y' = 0

5) is also 3*u_x' - 2*v_x' = 0

the only solution that satisfy 1,2,3,4,5 simultaniously is

u_x' = 0
u_y' = 0
v_x' = 0
v_y' = 0

which shows that f(z) is independent of x,y on all D and that f(z) is a constant.

 

[xaos]

now you need to finish this off. look at d/dz f(z) and analyticity.

 

[squaremeplz]

d/dz f(z) = u_x' + i*v_x = v_y' - i*u_y'

d/dz f(z) = 0 + i*0 = 0 - i*0

f(z) = c

is this ok?