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/
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