complex

purebladeknight · #1 $Old$ November 8th, 2009, 01:23 AM

express in trig form

$(-2 Ix)/(e^-Ix - e^(Ix)$ $- (8I) (e^(-Ix) + e^(Ix) (I/2) (e^(-Ix) - e^(Ix)$ $- (e^(-Ix) + (e^(Ix)/2 /(e^(-Ix) - e^(I x)^3$ $= 4/(e^(2 Ix) (e^(-Ix) - e^(Ix)^3 - (4 e^(2 Ix) / (e^(-Ix)$ $- e^(Ix)^3 + (12Ix) / (e^(-Ix) - e^(I x)^3 + (2Ix) / (e^(2Ix) (e^(-Ix)$ $- e^(I x)^3 + (2I) e^(2 Ix) x)/(e^(-Ix) - e^(Ix)^3$

sorry first time using latex. i dont know how to show the numerator over the denominator and the powers keep ending up looking larger than its supposed to be.

nvm figured it out

Prove It · #2 $Old$ November 8th, 2009, 01:27 AM

Quote:

Originally Posted by purebladeknight $View Post$

express in trig trig form

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Recall that $e^{i\theta} = \cos{\theta} + i\sin{\theta}$ .

You should be able to convert them all to sines and cosines, and then use trigonometric identities to simplify.

CaptainBlack · #3 $Old$ November 8th, 2009, 01:43 AM

Quote:

Originally Posted by purebladeknight $View Post$

express in trig form

$(-2 Ix)/(e^-Ix - e^(Ix)$ $- (8I) (e^(-Ix) + e^(Ix) (I/2) (e^(-Ix) - e^(Ix)$ $- (e^(-Ix) + (e^(Ix)/2 /(e^(-Ix) - e^(I x)^3$ $= 4/(e^(2 Ix) (e^(-Ix) - e^(Ix)^3 - (4 e^(2 Ix) / (e^(-Ix)$ $- e^(Ix)^3 + (12Ix) / (e^(-Ix) - e^(I x)^3 + (2Ix) / (e^(2Ix) (e^(-Ix)$ $- e^(I x)^3 + (2I) e^(2 Ix) x)/(e^(-Ix) - e^(Ix)^3$

sorry first time using latex. i dont know how to show the numerator over the denominator and the powers keep ending up looking large

Please revise this and make sure that your brackets match.

Also an exponent should be written e^{ix} in LaTeX.

Also^2 try asking WolframAlpha to simplify this.

CB