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Evaluating e^(i*pi) and i^i

Date: 07/25/99 at 08:16:52
From: julie balen
Subject: Investigating hyperbolic functions


I need to evaluate e^ipi and i^i. I know that e^ipi - 1 = 0, and that 
this uses the 5 most important constants in mathematics, but I do not 
really understand how or why it is used, or what i^i is. I read 
somewhere that i^i has an infinite number of solutions, but how can 
that be? And is that true?

Sorry to keep asking all these things, but they really interest me, 
and I need to know them for a project I am doing in school.

Thank you so much!

Date: 07/25/99 at 08:47:15
From: Doctor Jerry
Subject: Re: Investigating hyperbolic functions

Hi Julie,

Do you mean e^(i*pi)?  I'm easily confused, without parentheses. This 
can be done with the formula you gave, which is a special case of 
Euler's identity:

      e^(i*t) = cos(t) + i*sin(t).

If t = pi, then

     e^(i*pi) = cos(pi) + i*sin(pi) = -1.

Okay, that leaves the i^i question. The way to do these is to convert 
to polar coordinates and use logs. Let w be any complex number for 
which w = i^i. Writing w in polar form, which is r*e^(i*t), then

     ln(w) = ln(r) + i*t*ln(e)
           = ln(r) + i*t
           = i*ln(i)

Writing i in polar form,

             i = 1*e^(i*pi/2)

and so   ln(i) = i*pi/2.


     ln(r) + i*t = i(i*pi/2)
                 = -pi/2.

So,   t = 0   and   ln(r) = -pi/2. So, 

     w = r*e^(i*t)
       = e^(-pi/2) = 0.20787...

There are other answers. You can get them by doing the above, but 
start with

     w = r*e^(i*(t+k*2pi))

where k is any integer. I took k = 0. 

- Doctor Jerry, The Math Forum   
Associated Topics:
High School Imaginary/Complex Numbers

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