Thirteen Power of (1+i)

If i represents the imaginary unit, find real values for a and b such that (1+i)^{13}=a+bi.
Source: NCTM Mathematics Teacher, October 2006

Solution
(1+i)^2=1+2i+i^2
=2i
(1+i)^{13}=((1+i)^2)^6(1+i)
=(2i)^6(1+i)
=64\:i^6(1+i)
=64(-1)(1+i)
=-64-64i

Answer: a=-64,b=-64

Alternative solution
Write the complex number in polar form
1+i=\sqrt{2}\left (\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}\:i\right )
=\sqrt{2}\left (\text{cos} \dfrac{\pi}{4}+i\:\text{sin} \dfrac{\pi}{4}\right )
Applying De Moivre’s theorem
(1+i)^{13}=\left [\sqrt{2}\left (\text{cos}\dfrac{\pi}{4}+i\:\text{sin} \dfrac{\pi}{4}\right )\right ]^{13}
=(\sqrt{2})^{13}\left (\text{cos}\dfrac{13\pi}{4}+ i\:\text{sin}\dfrac{13\pi}{4}\right )
=64\sqrt{2}\left (\text{cos}\dfrac{5\pi}{4}+ i\:\text{sin}\dfrac{5\pi}{4}\right )
=64\sqrt{2}\left (-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}\:i\right )
=-64-64i

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About mvtrinh

Retired high school math teacher.
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