Hi fmunshim !!
A complex function f(z) can be expressed as:
f(z) = u(z) + i.v(z) = u(x,y) + i.v(x,y) = f(x,y)
where z = x + i.y and u,v are real functions of complex variable;
then dz = dx + i.dy;
If f(z) is analitical in a domain D then:
f(z+h) - f(z) f(x+j,y+k) - f(x,y)
f´(z) = lim -------------- = lim -----------------------
(h-->0) h (h-->0) h
where z = x + i.y and h = j + i.k;
If h is real:
f(z+h) - f(z) f(x+j,y) - f(x,y)
f´(z) = lim -------------- = lim --------------------- = df/dx(z)
(h-->0) h (j-->0) h
we will note df/dx(z) as fx(z).
If h is pure imaginary, h = i.k :
f(z+h) - f(z) f(x,y+k) - f(x,y)
f´(z) = lim -------------- = lim --------------------- =
(1/i).df/dy(z)
(h-->0) h (k-->0) i.k
= -i.fy(z)
The limit exists anyone is the way of approach to the zero, then
fx(z) = -i.fy(z) (eq.1)
As f(z) = u(z) + i.v(z) we have from eq.1:
fx = ux + i.vx = -i.fy = -i.(uy + i.vy) = vy - i.uy
then we have the Cauchy-Riemann Equations:
ux = vy and vx = -uy .
You can see other proof of this at "Cauchy-Riemann Equations -- from
MathWorld"
http://mathworld.wolfram.com/Cauchy-RiemannEquations.html
I will use this equations to proof your proposition:
a) Re f(z) = k.
f(z) = u(z) + i.v(z) with u(z) = k for all z in D.
then 0 = ux = vy and 0 = uy = -vx
then f´(z) = 0 for all z in D.
------------------------------
b) Im f(z) = k
f(z) = u(z) + i.v(z) with v(z) = k for all z in D.
then 0 = vy = ux and 0 = vx = -uy
then f´(z) = 0 for all z in D.
-------------------------------
c) |f(z)| = k
then |f(z)|^2 = k^2 = k.k ,
As |f(z)|^2 = u^2 + v^2 then if g(z) = |f(z)|^2,
and g(z) = j = k^2
gx = 2.(u.ux + v.vx) = 0 and gy = 2.(u.uy + v.vy) = 0, then
u.ux + v.vx = 0 and 0 = u.uy + v.vy = v.ux - u.vx ;
Solving this equations for ux and vx we have:
ux = vx = 0 unless u^2 + v^2 = 0 ,
Because g(z) is constant in D and g = u^2 + v^2, then if g = 0 for
some z in D,
then g(z) = 0 in D and follows that f(z) = 0 in D is constant.
Otherwise ux = vx = 0 , then uy = vy = 0 in D and f´(z) = 0 for all
z in D.
----------------------------------
Now we need to prove that if f´(z) = 0 for all z in a domain D, then
f(z) is constant in D.
First of all I will remember you the definitions of Domain and
Connected Set and an important property of the Domains.
Definition: A domain is a connected open set.
Definition: A connected set is a set which cannot be partitioned into
two nonempty subsets such that each subset has no points in common
with the closure of the other.
Property: In an open connected set of C or R^2 any two points of the
set can be joined by a path consisting of straight line segments (i.e.
a polygonal path), and we can set the polygonal path so that the lines
be parallel to the coordinates axis.
Now we can finish the proof:
f´(z) = ux(z) + i.vx(z), if this derivative is equal to zero then we
have
that all partial derivatives ux = vy , vx = -uy are all equal to zero
too.
Then u and v are constants along any straight line parallel to the
coordinates axis in the domain D.
As D is a open connected set then u(z) and v(z) are constants in D (it
follows from definition of continuity: we can approach to the point z
in D using any way), then
f(z) = u(z) + i.v(z) is constant in D.
I hope this helps you, if you have doubts, feel free to post a request
of clarification.
Thank you for use Google Answers.
Best Regards.
livioflores-ga |
Clarification of Answer by
livioflores-ga
on
10 Dec 2002 01:01 PST
Hi again fmunshi!!
I took the hardest way. I miss the "Open mapping property" part.
So you want that I use the Open mapping theorem that states:
"A nonconstant analytic function on a domain D is an open map. That is
a map which sends open sets to open sets."
Or in other words:
"If f(z) is a nonconstant analytic function in a domain D, the image
of D under the mapping w = f(z) is a domain in the w plane (i.e. open
connected sets map to open connected sets)."
a) Re f(z) = k for all z in D.
f(z) = k + i.v(z)
Then if w is in the W = (Image of D under f)= f(D) , w = k + i.v ,
where k is constant and v is in the image of the function v(z).
Now take w1 in W , then w1 = k + i.v1 ;
we have for every e>0, w = (k+d) + i.v1 (where 0 < d < e) is not in
W ,
then no exists e>0 so that all w that satisfied |w1-w| < e are in W ,
then W isn´t an open set, then like f is analitical by hypothesis, f
cannot be nonconstant, then f is constant.
-----------------------------------------
b) Im f(z) = k for all z in D.
f(z) = u(z) + i.k
Then if w is in the W = (Image of D under f)= f(D) , w = u + i.k ,
where k is constant and u is in the image of the function u(z).
Now take w1 in W , then w1 = u1 + i.k ;
we have for every e>0, w = u1 + i.(k+d) (where 0 < d < e) is not in
W ,
then no exists e>0 so that all w that satisfied |w1-w| < e are in W ,
then W isn´t an open set, then like f is analitical by hypothesis, f
cannot be nonconstant, then f is constant.
---------------------------------
c)|f(z)| = k for all z in D.
f(z) = u(x) + i.v(z)
Then if W = (Image of D under f)= f(D)
w is in W if and only if w = u + i.v (where u is in the image of u(z)
and v is in the image of v(z)) and u^2 + v^2 = k^2 .
Take w1 in W , w1 = u1 + i.v1 ,
for all e>0 we can take 0 < d < e, then
w = (u1+d) + i.v1 isn´t in W , in effect:
If u1 = 0 then w = d + i.v1 and v1^2 = k^2 ,
d^2 + v1^2 = d^2 + k^2 > k^2 then w isn´t in W.
If u1 > 0 then w = (u1+d) + i.v1 , then
(u1+d)^2 + v1^2 = u1^2 + 2.u1.d + d^2 + v1^2 = (k^2 + 2.u1.d + d^2) >
k^2
then w isn´t in W.
If u1 < 0 then w = (u1-d) + i.v1 , then
(u1-d)^2 + v1^2 = u1^2 - 2.u1.d + d^2 + v1^2 = (k^2 - 2.u1.d + d^2) >
k^2
then w isn´t in W.
Then we have that for all e > 0 exists w that satisfied |w1-w| < e
and w isn´t in W , then W isn´t an open set, then like f is analitical
by hypothesis, f cannot be nonconstant, then f is constant.
Please if you need some clarification please post a request for it,
and excuse me for the confusion.
Regards.
livioflores-ga
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