Real and Imaginary Part of Large Power

I came across this problem in a complex analysis book:

Find the real and imaginary parts of $(1 + i)^{100}$.

Now, this question is asked before polar form is introduced, so I am curious about alternate methods of solution. Of course, I could calculate this brute force (break it down by factoring $100 = 2 \cdot 2 \cdot 5 \cdot 5$), but that is tedious. As far as I know, there aren't any multiplicative properties for $Re$ and $Im$ either.

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What is $(1+i)^2$? What about $(1+i)^4$? Can you raise this to the power of 25? – user108903 Dec 16 '12 at 20:00
Has DeMoivre's Theorem been introduced? This would be a natural approach. – Clayton Dec 16 '12 at 20:00
@user108903 $(1 + i)^2 = 2i$ ... lesson learned. – tylerc0816 Dec 16 '12 at 20:02

An alternate method would be using the fact that $(1+i)^2 = 2i$. So then we have $(1+i)^{100} = (2i)^{50} = 2^{50}\cdot (i)^{50} = -2^{50}$.

For this case, the fact that $(1+i)^2=2i$ made this very easy to break down.

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The natural thing to do is to calculate a few small powers to see whether anything leaps out at you $-$ and in this case it does.

$(1+i)^2=2i$, so $(1+i)^4=(2i)^2=-4$, and $(1+i)^{100}=(-4)^{25}=-2^{50}$.

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Note that $(1+i)= \sqrt{2} \left(\dfrac1{\sqrt{2}} + i \dfrac1{\sqrt{2}} \right) = \sqrt{2} e^{i \pi/4}$. Hence, $$(1+i)^{100} = 2^{50} e^{i \pi/4 \times 100} = 2^{50} e^{i25 \pi} = 2^{50} e^{i \pi} = - 2^{50}$$

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