Re: Complex Analysis

In article <g7a2eh$bu8$1@xxxxxxxxxxxxxxxxx>,
T-Hope <s0456518@xxxxxxxxxxxx> wrote:

Thank you very much for the reply. I am still not with you, I'm sorry.
Here's how I see it:

Im=y
|
|
| this minus
| the Re axis
| \/
|
------------------------------------Re=x
* * * * * * * * |* * * * * * * * *
* * * * * * * * |* * * * * * * * *
* * * * * * * * |* * * * * * * * *
* * * * * * * * |* * * * * * * * *
* * * * * * * * |* * * * * * * * *
* * * * * * * * |* * * * * * * * *
* * * * * * * * |* * * * * * * * *



This maps to:

Im=y
|* * * * * * * * *
this minus |* * * * * * * * *
then Im axis >|* * * * * * * * *
|* * * * * * * * *
|* * * * * * * * *
|* * * * * * * * *
------------------------------------Re=x
|* * * * * * * * *
|* * * * * * * * *
|* * * * * * * * *
|* * * * * * * * *
|* * * * * * * * *
|* * * * * * * * *


No.

* - not in the set

So say z=exp([pi]i/6) \in S, then f(z)=exp([pi]i/3) which clearly
doesn't lie in the "complex plane minus the non-negative real axis".

That clearly doesn't lie in the set you drew above. But
it does lie in the complex plane minus the non-negative real
axis. Because (i) it's in the complex plane and (ii) it does
not lie on the non-negative real axis.

This is bugging me because the author simply brushes past this as if to
say it's obvious but something just isn't clicking. Thanks for your
patience.

-T

--
David C. Ullrich
.

Relevant Pages

  • Could you help me with complex numbers?
    ... math and try to study a bit more thn required by my school. ... But then it was defined that, in the complex plane, (Argand ... money it was say, distance, I could understand, since the real axis may ... Then the negative integers were ...
    (sci.math)
  • Re: Cartesian chart
    ... The axis contain measurement scales that intersect at 0. ... positive numbers lie above the horizontal axis (that ... Peltier Technical Services ...
    (microsoft.public.excel.charting)
  • Re: mandelbrot in imaginary space
    ... > i've extended the complex plane to 3 and 4 dimensions. ... > the unit of the imaginary axis is i. it is the the root of the negated ... > unit of the real axis. ...
    (sci.fractals)
  • Re: how can I return nothing?
    ... Tor Rustad said: ... point on the complex plane. ... I think you picture a single real axis. ... axis, as well as an imaginary axis for time. ...
    (comp.lang.c)
  • Re: polynomial question
    ... Long division will convince you. ... The solutions of this equation will lie on the unit circle in R^2 when ... thought of as the complex plane. ...
    (sci.math)