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"Compute $(1 + i)^{1000}$.

So far I have: $(1+i)^{4 (2^2 5^3)} $ but I am not sure how to proceed. Ideas?

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marked as duplicate by apnorton, RecklessReckoner, daw, Bookend, Steven Taschuk Jul 13 '14 at 17:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you tried computing $(1+i)^2$? – Mark Bennet Jul 13 '14 at 13:11
up vote 5 down vote accepted


${(i+1)}^{1000} = {({(i+1)}^{2})}^{500} = {(2i)}^{500} = {2}^{500}{i}^{500} = {2}^{500}{(i^2)}^{250} = {2}^{500}{(-1)}^{250}$

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why can you make the jump from $((1+i)^2)^{500} = (2i)^{500}$ ? – Ozera Jul 13 '14 at 13:18
@Ozera: ${(1+i)}^{2} = (1+i)(1+i) = 1^2+2i+i^2 = 1+2i-1 = 2i$ – barak manos Jul 13 '14 at 13:20
Shouldn't that $255$ be $250$? – Mark Dickinson Jul 13 '14 at 13:36
@MarkDickinson: Ooops :) Thanks – barak manos Jul 13 '14 at 13:40
@Ozera: Please note that in the original answer I wrote $255$ instead of $250$ (which essentially changes the answer from positive to negative). – barak manos Jul 13 '14 at 13:41

Powers of complex numbers are often easier in polar coordinates. The absolute value of $(1+i)$ is $\sqrt{2}$, and its angle is $\pi/4$. So the question becomes $$(\sqrt{2}e^{i\pi/4})^{1000}$$

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Here, $\exp(z)$ stands for $e^z$

$$\begin{align}(1+i)^{1000}&=(\sqrt{2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}))^{1000} \\ &= \sqrt2^{1000}\exp(1000\cdot\frac{\pi}{4}i)\\ &=\sqrt{2}^{1000}\exp(250\pi i) \\ &=2^{500}(\cos(250\pi)+ i\sin(250\pi)) \\ &\text{For even multiples of $\pi$, $\cos \theta =1 , \sin \theta = 0$} \\ &=2^{500}(1+0)=2^{500} \end{align}$$

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