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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.

  • 4
    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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    @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 $e^{2\pi i}=1$.