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How do I find $(1+i)^{100}$ without expanding $(1+i)$ 100 times?

Is there a quicker way to do this?

The hint was to find the modulus and argument of $1+i$ which I've got as $\sqrt{2}$ and $\pi/4$ but I'm not sure what to do from here.

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$$(1+i)^2=2i\qquad \Longrightarrow\qquad (1+i)^4=-4.$$

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  • $\begingroup$ Nice solution, it reduces the calculation directly to integers. $\endgroup$ – Martin Brandenburg Jun 26 '14 at 20:01
  • $\begingroup$ The simplest solution! Nice! $\endgroup$ – johannesvalks Jun 26 '14 at 20:04
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    $\begingroup$ No this solution distracts from the whole point of the question $\endgroup$ – John Fernley Jun 26 '14 at 22:17
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    $\begingroup$ @JohnFernley Another barometer problem? $\endgroup$ – Start wearing purple Jun 27 '14 at 8:44
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    $\begingroup$ Any difference from $(1+i)=\sqrt{2}e^{\frac{i\pi}{4}}$ in simplicity is very slight $\endgroup$ – John Fernley Jun 27 '14 at 13:47
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Hint: $$(1+i)=\sqrt{2}e^{\frac{i\pi}{4}}$$

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Hint:

$1+i=\sqrt{2} e^{i\pi/4}$

Therefore, $(1+i)^{100}= [\sqrt{2} e^{i\pi/4}]^{100}=\sqrt{2}^{100}e^{i100\pi/4}=[{{2^{0.5}}}]^{100}e^{i(24\pi+\pi)}=2^{50}e^{i\pi}=2^{50}(-1)=\boxed{-2^{50}}$

Your job: how did we get $1+i=\sqrt{2} e^{i\pi/4}$?

To do this kind of problem in general, look at the polar form of a complex number:

http://tutorial.math.lamar.edu/Extras/ComplexPrimer/Forms.aspx

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You can use De Moivre's formula, $1+i=\sqrt{2}(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4})$, so $(1+i)^{100}=(\sqrt{2})^{100}(\cos 25\pi+i \sin 25\pi)$

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The ( at least a ) key is that the powers are periodic (argument-wise) .One way of doing Complex multiplication is by doing it geometrically, you multiply two numbers by squaring their respective moduli and by adding their respective arguments. Notice that $1+i$ lies on the line $y=x$, so that its argument is $\pi/4$. What happens if you go around $\frac {2k\pi}{\pi/4}$ times? The key is that $\pi/4$ is "commensurate" to $2\pi$, meaning it divides exactly (an integer number of times) into $2 \pi$.

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