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### Complex Powers

```
Date: 09/28/2001 at 03:00:15
From: Bill Koppelman
Subject: Complex powers

How do I show that abs(z^i) < exp^pi where z is a complex number not
equal to 0?
```

```
Date: 09/28/2001 at 11:09:29
From: Doctor Mitteldorf
Subject: Re: Complex powers

Start by writing z as r*exp(it) where r and t are real, -pi<t<pi.
Then

z^i = r^i * exp(-t)

The first part is

r^i = exp(i*ln(r)) = cos(ln(r)) + i sin(ln(r))

This has a magnitude of 1. (Any real number raised to the i power has
magnitude 1.)

The second part is exp(-t), where t is between -pi and pi. The
greatest value this can have is exp(-(-pi)) = exp(pi).

So there's your proof.  Are you convinced?

I'm not.  Suppose I had begun by

...writing z as r*exp(it) where r and t are real, 0 < t < 2pi.

Then I would have finished by saying that the range of exp(-t) was
exp(-2pi) to exp(0), so the proof would now say abs(z^i) < 1.

Or I might have used the range 100pi < t < 102pi,

in which case I could have shown that abs(z^i) < exp(-100).

What goes on here? I encourage you to let this paradox rumble around
resolve it. Meanwhile, take a look at this page from our archives:

e^(i*pi) = -1: pi = 0 ?
http://mathforum.org/dr.math/problems/koehler10.17.97.html

- Doctor Mitteldorf, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Imaginary/Complex Numbers
High School Number Theory
High School Transcendental Numbers

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