Dec 16, 2012 Thread starter #1 P Poly Member Nov 26, 2012 32 How do I find the imaginary part of $\displaystyle \frac{1}{i}xe^{-ix}+e^{ix}$? Last edited: Dec 16, 2012
Dec 16, 2012 Admin #2 M MarkFL Administrator Staff member Feb 24, 2012 13,775 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: $\displaystyle e^{\theta i}=\cos(\theta)+i\sin(\theta)$
Dec 16, 2012 Thread starter #3 P Poly Member Nov 26, 2012 32 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.
