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Solving x^x = i


Date: 12/23/2000 at 01:04:03
From: Guy Pledger
Subject: Equation for x^x=i

This might be a nebulous question, but is there a way to find what 
number raised to it's own power would yield i, that is, x^x = i? I 
understand how imaginary exponents work and have done a lot with them, 
but I see no way to solve this. Can you help?

Thanks,
Guy Pledger


Date: 12/23/2000 at 06:32:29
From: Doctor Mitteldorf
Subject: Re: Equation for x^x=i

Dear Guy,

It's not nebulous at all - yours is a very well-formed and specific 
question. I'd go about it using the Euler formula:

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

Start by noting that if e^(it) = i, then t = pi/2. Therefore,

     ln(i) = i*pi/2

Now take your equation, and take the natural log of both sides:

     x*ln(x) = i*pi/2

Let's write x as r*e^(it), where both r and t are real. Then we have:

     ln(x) = ln(r) + it

     r*e^(it) * (ln(r)+it) = i*pi/2

We can break this equation into two real equations, identifying the 
real and imaginary parts of both sides. (The real part of the right 
side is zero.)

        Real part gives:   r*cos(t)*ln(r) - r*sin(t)*t = 0
   Imaginary part gives:   r*sin(t)*ln(r) + r*cos(t)*t = pi/2

You can divide the first equation through by r, and solve to get:

     ln(r) = t*tan(t)
         r = e^(t*tan(t))

Substituting this into the second equation, we have a single equation 
for t:

     t*e^(t*tan(t)) * [sin(t)*tan(t) + cos(t)] = pi/2

This equation must be solved numerically, and it gives approximately 
t = 0.68845. The first equation then gives r = 1.7619. Hence:

     x = r*e^(it)= 1.3606 + i*1.1194

I'll leave it to you to take x^x and check that it indeed comes out 
close to i.

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

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