- Sep 10, 2004
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I'm trying to fully understand why:
The answer should be... pi^-1 * (cos[2ln(pi)]) + pi^-1 * (i * sin[2ln(pi)])
- I do know that you can separate the pi^-1 and put it in the form pi^-1 * (cos mx + i * sinmx)
- The two comes from the m, which is 2.
Why is it ln(pi) though?... if you have pi^-1 * pi^(m*ln(cosx+isinx)), how do you convert that into the answer?
Solved: Thanks dullard!
Use euler's formula (e^ix = cosx + isinx) to put pi^(-1+2i) in the form a+ib
The answer should be... pi^-1 * (cos[2ln(pi)]) + pi^-1 * (i * sin[2ln(pi)])
- I do know that you can separate the pi^-1 and put it in the form pi^-1 * (cos mx + i * sinmx)
- The two comes from the m, which is 2.
Why is it ln(pi) though?... if you have pi^-1 * pi^(m*ln(cosx+isinx)), how do you convert that into the answer?
Solved: Thanks dullard!
