Poly1 Messages 32 Reaction score 0 Thread starter Dec 16, 2012 #1 How do I find the imaginary part of $\displaystyle \frac{1}{i}xe^{-ix}+e^{ix}$? Last edited: Dec 16, 2012
MarkFL Gold Member MHB Messages 13,284 Reaction score 12 Dec 16, 2012 #2 For the first term, I would rewrite i with a negative exponent, then apply: $\displaystyle i^{n}=i^{n+4k}$ where $\displaystyle k\in\mathbb{Z}$ For the second term, apply Euler's formula: $\displaystyle e^{\theta i}=\cos(\theta)+i\sin(\theta)$
For the first term, I would rewrite i with a negative exponent, then apply: $\displaystyle i^{n}=i^{n+4k}$ where $\displaystyle k\in\mathbb{Z}$ For the second term, apply Euler's formula: $\displaystyle e^{\theta i}=\cos(\theta)+i\sin(\theta)$
Poly1 Messages 32 Reaction score 0 Dec 16, 2012 #3 Sorry there was an $i$ missing from the first part. But I got the answer using your suggestion. Thanks.
Sorry there was an $i$ missing from the first part. But I got the answer using your suggestion. Thanks.