The ith Root of -1
Date: 09/16/1999 at 15:54:01 From: Carl Ford Subject: Imaginary roots I would like to know how the ith root of -1 (or (-1)^(1/i)) equals 23.14069... I know that this works and why, but not how. It works by rearranging the equation e^(Pi*i)+1 = 0 to e^Pi = (-1)^(1/i), and e^Pi = 23.14069... but I can't understand how it works. It makes no sense to me that: ___ _i / \/ -1 = 23.14069... but ___ _i / \/ 1 = 1 Thanks, Carl
Date: 09/16/1999 at 22:23:49 From: Doctor Schwa Subject: Re: Imaginary roots Carl, This is an excellent question! It certainly seems ridiculous that taking a root of a unit-sized number could suddenly be so big, even if the root is complex. Here's the key point: The expression e^(pi*i) is also the same as e^(-pi*i) because the pi here is referring to an angle in radians, so you can add or subtract any multiple of 2pi from it without changing the value. Thus (-1)^(1/i) is equal to e^pi but it's ALSO equal to e^-pi and ALSO to e^(-3pi) and e^(-5pi) and e^(37pi) and ... Once you allow complex numbers into your exponents, logarithms become many-valued things, so roots (which can be defined in terms of logarithms) also have many possible answers. I hope that helps clear things up. If you'd like to know more about e^(a + bi) in general, or roots or logarithms of complex numbers, please do write back with more questions. - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/
Date: 09/22/1999 at 14:53:47 From: The Professor Subject: Re: Imaginary roots I don't really understand the above answer. Could you please show me how you would calculate the ith root of -1 if I didn't know that it equaled e^pi? My graphing calculator can take the ith root of -1, and I can't figure out how it does that. Also, how would you take the ith root of anything else, like the ith root of 9? (My calculator does that too, but I can't figure it out either.)
Date: 09/22/1999 at 15:51:33 From: Doctor Anthony Subject: Re: Imaginary roots We can write -1 = cos(pi) + i.sin(pi) -1 = e^(i.pi) Take ith root of both sides (-1)^(1/i) = [e^(i.pi)]^(1/i) = e^[(i.pi)/i] = e^(pi) = 23.14069... There are of course multiple answers because we could have started with -1 = e^(i.3pi) or e^(i.5pi) or ... - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/
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