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Just a basic question. I have $e^{-i9\pi/4}$, and I'm struggling to understand why this is equal to $e^{-i\pi/4}$.

Thanks in advance for any help.

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    Poles? $ $ $ $ $ $2011-05-23

2 Answers 2

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Because

  • $e^{i\theta}=\cos\theta + i\cdot \sin\theta$,

  • $e^{-i\theta}=\cos(-\theta)+i\cdot \sin(-\theta)=\cos\theta - i \sin(\theta)$.

  • $\sin(2\pi+t)=\sin(t)$, same for cosine also.

  • $\sin\left(\frac{9\pi}{4}\right)=\sin \left(2\pi + \frac{\pi}{4}\right)$.

So that \begin{align*} e^{-i\frac{9\pi}{4}} &=\cos\frac{-9\pi}{4}+i \sin\frac{-9\pi}{4}\\ &=\cos\frac{9\pi}{4} +i \sin\frac{-9\pi}{4}\\ &=\cos\frac{9\pi}{4} -i\sin\frac{9\pi}{4}\\ &= \cos\left(2\pi + \frac{\pi}{4}\right)-i\sin\left(2\pi +\frac{\pi}{4}\right)\\ &=\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\\ &=e^{-i\frac{\pi}{4}} \end{align*}

Also while doing such problems, these formulae may come handy:

  • $\sin(\pi+\theta)= - \sin\theta\ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

  • $\sin(\pi-\theta) = \sin\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

  • $\sin\bigl(\frac{\pi}{2}+\theta\bigr) = \cos\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

  • $\sin\bigl(\frac{\pi}{2} -\theta\bigr) = \cos\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

  • $\cos(\pi+\theta)=-\cos\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

  • $\cos(\pi - \theta)=-\cos\theta\ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

  • $\cos\bigl(\frac{\pi}{2}-\theta\bigr)=\sin\theta \ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

  • $\cos\bigl(\frac{\pi}{2} + \theta\bigr)= -\sin\theta\ ; \qquad 0 \leq \theta \leq \frac{\pi}{2}$

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    So is $e^{-i5\pi/4}$ also equal to $e^{-i\pi/4}$, because $5 \equiv 1 \ (\text{mod} \ 4)$?2011-05-23
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    @Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake.2011-05-23
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    So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well.2011-05-23
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    @Doodle: For that you have to see where the Sines and cosines are positive /negative. So you have $5\pi/4 = \pi + \pi(4)$. So $\sin$ becomes negative and cos becomes positive.2011-05-23
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    @Doodle: $e^{-5\pi/4}= \cos(-5\pi/4) +i\sin(-5\pi/4) =\cos(5\pi/4)-i\sin(5\pi/4) = \cos\frac{\pi}{4} +i\sin\frac{\pi}{4}$2011-05-23
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    @Chandru: $\cos(5\pi/4)\neq \cos(\pi/4)$.2011-05-23
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    @Jonas: Oops again a blemish :(2011-05-23
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    @Jonas: Yeah it is $-\cos\frac{\pi}{4}$2011-05-23
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    @Jonas: Thought it was in the fourth quadrant while typing2011-05-23
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Note that $9\pi/4 = (1+8)\left(\pi/4\right) = \pi/4 + 2\pi$ and $e^{2\pi i}=1$.