# Complex exponential function

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?    –  Did May 23 '11 at 14:10

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 $\cos$ also.

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

\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\bigl(2\pi + \frac{\pi}{4}\bigr)-i\sin\bigl(2\pi +\frac{\pi}{4}\bigr) \\ &=\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}=e^{-i\frac{\pi}{4}} \end{align*}

Also while doing such problems, these formula's 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 $5 \equiv 1 \ (\text{mod} \ 4)$? –  Jonas Meyer May 23 '11 at 14:17
@Jonas: Sorry. I made a mistake. Then i went to eat, when i realized that i had made a mistake. –  user9413 May 23 '11 at 14:41
So what would e−i5π/4 equal? I thought that would be e−iπ/4 as well. –  Dooodle May 23 '11 at 14:44
@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. –  user9413 May 23 '11 at 14:47
@Doodle: $e^{-5\pi/4}= \cos(-5\pi/4) +i\sin(-5\pi/4) =\cos(5\pi/4)-i\sin(5\pi/4) = \cos\frac{\pi}{4} +i\sin\frac{\pi}{4}$ –  user9413 May 23 '11 at 14:49

Note that $9\pi/4 = (1+8)\left(\pi/4\right) = \pi/4 + 2\pi$ and $e^{2\pi i}=1$.

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