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### 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

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
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!

lots of topics throughout your study of complex analysis.

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Analysis
College Imaginary/Complex Numbers

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