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| Author | Message / Information |
| elliotchance Quote | Reply | | Complex power problem posted on: 5/25/2006 6:03:26 AM This problem has been bugging me for ages, how do you evaluate e^i(pi/18), *without* using a calculator Thanks |
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Euler
Quote | Reply | |
Complex power problem
replied on: 6/4/2006 11:39:05 PM I know that a famous result of Leonhard Euler is that: e^Ci=cos(Cx)+isin(Cx), where C is any constant (such as pi/18 in this case) As for evaluating, with or without a calcultor, this is impossible, as i has no quantifiable value. Hope this helps! |
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Dante77
Quote | Reply | This message was updated on 8/25/2006 2:08:39 PM by Dante77 |
Complex power problem
replied on: 8/25/2006 2:07:45 PM Since it is known how to solve cubics directly, I'm only going to outline a method [cheap way out lol]. As you already know Cos[3x] + I*Sin[3x] = ( Cos[x] + I*Sin[x] ) ^3 By expanding the right handside, and comparing real and imaginary terms on each side of the equation we have. Cos[3x] = Cos^3[x]-3Sin^2[x]Cos[x] [equation1] Sin[3x] = 3Sin[x]- 4Sin^3[x] [equation2] The idea is to plug in x = pi/6 and solve equations 1 and 2 for Cos[x] and Sin[x], noting Sin[pi/6]= 0.5 and Cos[pi/6]=Sqrt[3]/2 Its easier to start with [equation2] because it only uses Sine terms. All you need to do now is research how to solve Cubic equations! Hope this helps |
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Euler
Quote | Reply | |
Complex power problem
replied on: 11/1/2006 12:19:37 AM How exactly did you arrive at the conclusion that cos(3x)+isin(3x)=(cos(x)+isin(x))^3?? Further more, where did the 3x and the x come from?? SHouldn't it be cos(pi/18)+isin(pi/18)?? |
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