Re: Complex Analysis - polar form notation!?!?!?!

On Feb 25, 3:49 pm, "Joshua Lindman" <i.love.jeevi...@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@)

So z = r e^{i @} is like an intermediate function transforming
coordinates (r,@) to (x,y) , correct?

More like transforming (x,y) to (r,@) [by mapping (r,@) to (x,y)]

.

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