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### Logs of Complex Numbers

```Date: 02/11/2004 at 22:27:22
From: Brittany
Subject: Log with complex numbers

Give an example showing that Log(z1/z2) does not always equal Log(z1)
- Log(z2) where z1 and z2 are complex numbers.  Each time I try to
plug in different complex numbers, such as z1 = e^i*pi/3 and
z2 = e^2i*pi/3, I find that the two calculations are equal.  I can't
find a counterexample.

```

```
Date: 02/12/2004 at 09:28:33
From: Doctor Vogler
Subject: Re: Log with complex numbers

Hi Brittany,

The complex logarithm is an interesting function.  Simpler complex
functions are defined everywhere except at certain points called
"poles" where they behave like divisions by zero.  But the logarithm
has a thing called a "branch" where you have to make a jump.  Because
the function f(z) = e^z has the property

f(z + 2*pi*i) = f(z),

there isn't only one inverse (or logarithm).  Well, we pick one of
those inverses for the function Log, but it causes there to be a jump.

Think of it this way.  If you make a 3-D graph with the complex plane
on the x-y coordinates and the value of the imaginary part of the
logarithm on the z coordinate, then what we really *should* get is a
spiral that is flat when you go straight out from the center x = y = 0
but winds upwards as you go counterclockwise around the center, and
increases by 2*pi each time around.

But that's not a function because a function can only have one value,
so we limit the domain by just taking one loop.  That means the spiral
starts on the negative x axis at z = -pi, does one counterclockwise
loop and ends on the same negative x axis but higher, at z = pi.  So
there is this break where the complex log jumps by 2*pi along the
negative x axis.

So in order to make Log (z1/z2) different from Log(z1) - Log(z2), you
just have to go over this break.  In other words, find a Log(z1) whose
imaginary part is negative near -pi and a Log(z2) whose imaginary part
is positive near pi (or vice-verse), and then subtracting them will be
on some other part of the logarithm spiral that was cut off in order
to make Log a function.

If you have any questions or need more help, please write back and
show me what you have been able to do, and I will try to offer further
suggestions.

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

```
Date: 02/12/2004 at 12:16:43
From: Brittany
Subject: Log with complex numbers

I tried z1 = e^i5pi/3 and z2 = e^i*pi/3 so that Log(z1) = -i*pi/3 and
Log(z2) = i*pi/3, and when you subtract these, you get -i2pi/3, which
is what I get when I calculate Log(z1/z2).  Am I not looking at the
correct z1 and z2?
```

```
Date: 02/12/2004 at 12:42:29
From: Doctor Vogler
Subject: Re: Log with complex numbers

Hi Brittany,

You need imaginary parts closer to pi and -pi.  Try making

Log(z1) = -i*2*pi/3
Log(z2) = i*2*pi/3

If you have any questions or need more help, please write back.

- Doctor Vogler, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Imaginary/Complex Numbers
High School Functions
High School Imaginary/Complex Numbers
High School Logs

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