Complex Analysis Geometry Proof

Date: 09/28/2004 at 14:14:42
From: LW
Subject: Complex Number Geometry Question

I'm looking for a proof that arg z1 + arg z2 = arg(z1z2).  I have a
good guideline to do this backwards [arg(z1z2) = z1 + arg z2], but the
proof that I have at this point is very likely to be incorrect.  I'd
appreciate your help a lot.

I have a hard time distinguishing how the guidelines for the second 
variation could be different from the first one.

I want to assume that Betta (Q) angle is in the [arg z1 + arg z2].
Can this be OK?  I want to say that Q1*Q2 is in [arg z1 + arg z2], but
that looks like a bunch of garbage to me!

Date: 09/28/2004 at 14:18:32
From: Doctor Roy
Subject: Re: Complex Number Geometry Question

Hi LW,

Thanks for writing to Dr. Math.

The easiest way to prove the statement is by using the phasor form of
complex numbers:

    z = abs(z)*exp(i*arg(z))

   z1 = abs(z1)*exp(i*arg(z1))
   z2 = abs(z2)*exp(i*arg(z2))

Does this help?  Please feel free to write back with any questions you
may have.

- Doctor Roy, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/28/2004 at 18:19:00
From: LW
Subject: Complex Number Geometry Question

I just started, as of yesterday, reading an introduction to Arg and 
arg, so basing my proof on use of the phasor form for z won't make my 
professor very happy.  I need to use Betta (Q) angles to do it.  For 
example Q = Q1+Q2.  Then it would be something like; since B1 is in 
the set z1, and Q2 is in the set z2, then Q is in the set of 
arg(z1z2) ...

I worked on it more, but I don't think I have a robust proof yet.  If 
you have any other thoughts on it, please respond--I appreciate your 
help a lot.

Date: 09/29/2004 at 15:38:27
From: Doctor Schwa
Subject: Re: Complex Number Geometry Question

Maybe a proof like the one found in the first chapter of Tristan
Needham's book, Visual Complex Analysis, is the sort of thing you have
in mind?  I recommend the book VERY highly.

The proof there is all based on similar triangles.

The idea is to say that i * z2 is a 90 degree rotated copy of z2, so 
if z1 = a+bi, then (a+bi)z2 corresponds to a triangle formed by (a 
copies of z2) and (b copies of a 90 degree rotated z2).  Then the 
vector addition pretty quickly shows that the endpoint of that is 
going to have an angle of arg z1 + arg z2.

Is that the kind of proof you want?

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 09/29/2004 at 17:06:48
From: LW
Subject: Thank you (Complex Number Geometry Question)

Yes, this proof goes along very well with what I'm doing in this 
class.  It is also easy to understand.  Thank you for it!

Date: 09/29/2004 at 20:07:37
From: Doctor Schwa
Subject: Re: Thank you (Complex Number Geometry Question)

Don't thank me, thank professor Needham!

Better yet, go buy his book!  It's really great and will help you with
lots of topics throughout your study of complex analysis.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/