# complex

• Nov 7th 2009, 11:23 PM
complex
express in trig form

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

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
• Nov 7th 2009, 11:27 PM
Prove It
Quote:

express in trig trig form

$\displaystyle ((-2 I) x)/(E^((-I) x) - E^(I x)) - ((8 I) (E^((-I) x) + E^(I x)) ((I/2) (E^((-I) x) - E^(I x)) - ((E^((-I) x) + E^(I x)) x)/2))/(E^((-I) x) - E^(I x))^3 == 4/(E^((2 I) x) (E^((-I) x) - E^(I x))^3) - (4 E^((2 I) x))/(E^((-I) x) - E^(I x))^3 + ((12 I) x)/(E^((-I) x) - E^(I x))^3 + ((2 I) x)/(E^((2 I) x) (E^((-I) x) - E^(I x))^3) + ((2 I) E^((2 I) x) x)/(E^((-I) x) - E^(I x))^3$

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

You should be able to convert them all to sines and cosines, and then use trigonometric identities to simplify.
• Nov 7th 2009, 11:43 PM
CaptainBlack
Quote:

express in trig form

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

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

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

Also^2 try asking WolframAlpha to simplify this.

CB