Re: Complex Analysis - polar form notation!?!?!?!
- From: Narcoleptic Insomniac <i_have_narcoleptic_insomnia@xxxxxxxxx>
- Date: Sun, 25 Feb 2007 17:39:58 EST
On Feb 25, 2007 2:49 PM CT, i.love.jeevitha@xxxxxxxxx wrote:
On Feb 24, 7:57 pm, Narcoleptic Insomniac
<i_have_narcoleptic_insom...@xxxxxxxxx> wrote:
On Feb 24, 2007 2:56 PM CT, i.love.jeevi...@xxxxxxxxx wrote:
My complex analysis book says...
(u,v,x,y in |R)
Suppose w = u + i*v is the value of the complex
function " f " at z = x + i*y
i.e., w = u + i*v = f(x + i*y)
Each of the real numbers u and v depends on the
real variables x and y, and it follows that f(z)
can be expressed in terms of a pair of real-valued
functions of two real variables x, y:
w = f(z) = u(x,y) + i*v(x,y)
***
I stared at this for a while, dazed and a little
confused by the notation, then I fell back on
the "complex number is just an ordered pair of
reals" idea. So I thought of this last equation as
just something like:
(u,v) = f(x, y) = ( u(x,y) , v(x,y) )
Then it made sense to me. But after this the book
says...
***
If the polar coordinates r and @ (theta) are used
instead of x and y, then
u + i*v = f( r*exp( i*@) ) where w = u + i*v and
z = r*exp( i*@). In that case we may write
f(z) = u(r, @) + i*v(r, @)
***
Other than laughing at the use of "w" for no
reason whatsoever, I was a bit confused by what this
notation really meant (I had a feeling but I
couldn't pinpoint it).
In the x,y coordinate case, since z= (x,y),
nothing special was being done by splitting the
equation up into two real valued parts. But here
I don't know what's going on.
I think the polar form case is drastically
different than the x,y case. We're essentially
using a composite function here, no?
z = r*exp(i*@) would be an intermediate function
mapping r, @ coordinates to x,y coordinates.
No, there need not be such an intermediate mapping.
In fact, I believe your troubles are comming from this
very notion.
But throughout the book so far the author treats
polar form as something very natural, as if it's
just "substitute r*exp( i*@) for z and everything
works." No mention of composite functions.
Why do you think it is so unnatural?
We're not 'substituting' r e^{i@} for z - we are
just using the polar coordinate system instead of the
cartesian system.
Perhaps you were mislead by the statement that z =
x + iy. Usually z is just taken to mean any arbitrary
complex number - the properties of the field of
complex numbers are independent of the coordinate
system we choose.
Can anyone make sense of this f(z) = u(r, @) +
i*v(r, @) ?
The expression makes some sense but what is it
really? We have a function of the variable z, which
really is a (x,y) equal to a two functions of
(r, @), but not just that, one of them is
multiplied by i.
So this translates to (since w = (u,v) and z =
(x,y)):
(u,v) = f(x,y) = ( u(r, @) , v( r, @) ) ?
This make no sense whatsoever to me. Someone
please help.
Actually, the more I read your post, the more
confused I get!
What you have is a mapping f:C --> C. Instead of
using cartesian coordinates in the domain, we are
going to use polar coordinates.
Each point z in the domain can be represented by
pair (r, @) by letting z = r e^{i @}. There is no
need to begin with (x, y) and somehow obtain (r, @)
- we are *starting* with (r, @).
When we write z = r e^{i @}
We are mapping (r,@) to (x,y) because
z = r e^{i @} = r(cos@ + i sin@) = rcos@ + i rsin@ =
(rcos@, rsin@)
No, when we write z = r e^{i @} we are attaching the
ordered pair (r, @) to the element z in the complex plane.
In other words, we are interpreting the field of complex
numbers as the set of ordered pairs
C = {(r, @) : r >= 0, 0 <= @ < 2pi}
...that satisfy certian algebraic properties.
So z = r e^{i @} is like an intermediate function
transforming coordinates (r,@) to (x,y) , correct?
No, but the mapping f:C --> C with
f(r, @) = f(r e^{i@}) = (r cos(@), r sin(@))
...that you gave above is.
I believe someone earlier suggested that you experiment
with the mapping g:C --> C such that g(z) = z^2; let's
consider that for a moment. We are going to use polar
coordinates in our domain and cartesian in our codomain.
This means that we are going to BEGIN with z = r e^{i@} -
WE'RE ASSOCIATING EVERY ELEMENT z WITH SOME ORDERED
PAIR (r, @) - and we're going to associate every element
f(z) = u + iv in the image with the ordered pair (u, v).
Now g(z) = z^2 = r^2 e^{2i@} = r^2 [cos(2@) + i sin(2@)]
...so...
g(r, @) = (r^2 cos(2@), r^2 sin(2@))
...which implies...
g(r, @) = u(r, @) + i v(r, @)
...where...
u(r, @) = r^2 cos(2@) ...and... v(r, @) = r^2 sin(2@).
There are two important things that should be noted here.
First, there was no need for an intermediate mapping
f(r, @) = r cos(@) + i r sin(@); we simply use the
identity e^{it} = cos(t) + i sin(t).
Sure, we *could* consider this identity as a mapping on
it's own right (and it's a very important one), but there
is no *need* to. For if we did, the composition would
look something like this.
(g o f)(z) = (g o f)(r, @) = (g o f)(r e^{i@})
= g(r cos(@), r sin(@))
= g(r cos(@) + i r sin(@))
= [r cos(@) + i r sin(@)]^2
= r^2 cos(@)^2 + 2i r cos(@) sin(@) - r^2 sin(@)^2.
The other thing that should be noted is the geometric
interpretation of the mapping z |--> z^2.
Recall that we had g(r, @) = r^2 e^{2i@}. If we consider
the image in polar coordinates instead of cartesian we
get g(r, @) = (r^2, 2@). This implies that the mapping
will square the length of a vector and double the
argument.
Regards,
Kyle Czarnecki
In your explanation of
f(z) = f(r e^{i @}) = u(r,@) + i v(r,@)
Even the form of the value is a complex number of
form u + i v
say (x,y) = g(r,@) = (rcos@, rsin@) -same as- r e^{i@}
*ignore the actual details of g(r,@) for now,
just assume a function exists that can do this
mapping*
and our function is f(x,y) = ( h(x,y) , k(x,y) )
so,
(u,v) = f(g(r,@)) = ( h(g(r,@)) , k(g(r,@)) )
so u = h(g(r,@))
v = k(g(r,@))
So it is in fact an intermediate function z=g(r,@)
at work. Am I right?
(see above)
.
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