Complex analysis with constant.
- From: "mina_world" <mina_world@xxxxxxxxxxx>
- Date: Fri, 1 Jun 2007 23:08:50 +0900
Hello sir~
If f'(z) = 0 everywhere in a domain D,
then f(z) must be constant throughout D.
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We write f(z) = u(x,y) + i.v(x,y).
Then, assuming that f'(z) = 0 in D,
we note that u_x + i.v_x = 0.
and, in view of the Cauchy-Riemann equations, v_y - i.u_y = 0.
Consequnetly, u_x = u_y = v_x = v_y = 0 at each point in D.
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is this not really sufficient ?
My book append to this proof.
Namely,
Now u_x and u_y are the x and y components of the vector grad u,
and the component of grad u at a given point in a specific direction
is the directional derivative of u there in that direction.
So the fact that u_x and u_y are always zero means that grad u
is always the zero vector and hence that any directional derivative of u
is zero.
[Namely,
Each pair of points P, P' in D can be joined by a polygonal line,
consisting of a finite number of line segments joined end to end,
that lies entirely in D.
This line called L.(from P to P')
s : length over L from P.
U : unit vector over L with s.
directional derivative of u : du / ds = (grad u).U = (u_x, u_y).U = 0.
so, u is constant over L.]
Consequently, u is constant along any line segment lying entirely in D;
and, since there is always a finite number of such line segment,
joined end to end, connecting any two points in D,
the values of u at those points must be the same.
We may conclude, then, that there is a real constant a such that
u(x,y) = a throughout D.
Similarly, v(x,y) = b; and it follows that f(z) = a+bi at each point in D.
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is this really need ?
.
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