Treba mi postupak.

15. Vrednost izraza [inlmath]\left(1+i\right)^{2012}+\left(1-i\right)^{2012}[/inlmath] je:

[inlmath]A)\;2^{1007}\quad[/inlmath] [inlmath]\enclose{circle}{B)}\;-2^{1007}\quad[/inlmath] [inlmath]C)\;2^{2012}i\quad[/inlmath] [inlmath]D)\;-2^{1007}i\quad[/inlmath] [inlmath]E)\;2^{2012}\quad[/inlmath] [inlmath]N)\;\mbox{ne znam}[/inlmath]

[dispmath]\left(1+i\right)^{2012}+\left(1-i\right)^{2012}[/dispmath]
Posmatramo prvih nekoliko stepena kako broja [inlmath]\left(1+i\right)[/inlmath], tako i broja [inlmath]\left(1-i\right)[/inlmath] i pokušamo da uočimo pravilnost:
[dispmath]\left(1+i\right)^2=1+2i+i^2=\cancel 1+2i-\cancel 1=2i[/dispmath][dispmath]\left(1+i\right)^4=\left[\left(1+i\right)^2\right]^2=\left(2i\right)^2[/dispmath][dispmath]\left(1+i\right)^6=\left[\left(1+i\right)^2\right]^3=\left(2i\right)^3[/dispmath][dispmath]\left(1+i\right)^8=\left[\left(1+i\right)^2\right]^4=\left(2i\right)^4[/dispmath][dispmath]\cdots[/dispmath][dispmath]\left(1+i\right)^{2k}=\left(2i\right)^k[/dispmath]
Sličnu pravilnost uočavamo i za [inlmath]\left(1-i\right)[/inlmath]:
[dispmath]\left(1-i\right)^2=1-2i+i^2=\cancel 1-2i-\cancel 1=-2i[/dispmath][dispmath]\left(1-i\right)^4=\left[\left(1-i\right)^2\right]^2=\left(-2i\right)^2[/dispmath][dispmath]\left(1-i\right)^6=\left[\left(1-i\right)^2\right]^3=\left(-2i\right)^3[/dispmath][dispmath]\left(1-i\right)^8=\left[\left(1-i\right)^2\right]^4=\left(-2i\right)^4[/dispmath][dispmath]\cdots[/dispmath][dispmath]\left(1-i\right)^{2k}=\left(-2i\right)^k[/dispmath]
[dispmath]\left(1+i\right)^{2012}+\left(1-i\right)^{2012}=\left(2i\right)^{1006}+\left(-2i\right)^{1006}=\left(2i\right)^{1006}+\cancelto{1}{\left(-1\right)^{1006}}\cdot\left(2i\right)^{1006}=[/dispmath][dispmath]=2\cdot\left(2i\right)^{1006}=2\cdot 2^{1006}\cdot i^{1006}=2^{1007}\cdot i^{4\cdot 251+2}=2^{1007}\cdot i^{4\cdot 251}\cdot i^2=[/dispmath][dispmath]=2^{1007}\cdot\left(i^4\right)^{251}\cdot\left(-1\right)=-2^{1007}\cdot 1^{254}=-2^{1007}[/dispmath]

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Može i preko trigonometrijskog oblika:
Prvo transformišemo u trigonometrijski oblik broj [inlmath]\left(1+i\right)[/inlmath]:
[dispmath]\rho=\sqrt{1^2+1^2}=\sqrt 2[/dispmath][dispmath]\varphi=\mathrm{arctg}\:\frac{1}{1}=\frac{\pi}{4}+2k\pi,\quad k\in\mathrm{Z}[/dispmath][dispmath]1+i=\sqrt 2\left[\cos\left(\frac{\pi}{4}+2k\pi\right)+i\sin\left(\frac{\pi}{4}+2k\pi\right)\right][/dispmath]
Isto tako i [inlmath]\left(1-i\right)[/inlmath]:
[dispmath]\rho=\sqrt{1^2+1^2}=\sqrt 2[/dispmath][dispmath]\varphi=\mathrm{arctg}\:\frac{-1}{1}=-\frac{\pi}{4}+2k\pi,\quad k\in\mathrm{Z}[/dispmath][dispmath]1-i=\sqrt 2\left[\cos\left(-\frac{\pi}{4}+2k\pi\right)+i\sin\left(-\frac{\pi}{4}+2k\pi\right)\right][/dispmath][dispmath]1-i=\sqrt 2\left[\cos\left(\frac{\pi}{4}+2k\pi\right)-i\sin\left(\frac{\pi}{4}+2k\pi\right)\right][/dispmath]
[dispmath]\left(1+i\right)^{2012}+\left(1-i\right)^{2012}=[/dispmath][dispmath]=\left(\sqrt 2\right)^{2012}\left[\cos\left(\frac{\pi}{4}+2k\pi\right)+i\sin\left(\frac{\pi}{4}+2k\pi\right)\right]^{2012}+\left(\sqrt 2\right)^{2012}\left[\cos\left(\frac{\pi}{4}+2k\pi\right)-i\sin\left(\frac{\pi}{4}+2k\pi\right)\right]^{2012}=[/dispmath]
primenimo Muavrovu formulu
[dispmath]=2^{1006}\left[\cos\left(\frac{2012\pi}{4}+2012\cdot 2k\pi\right)+\cancel{i\sin\left(\frac{2012\pi}{4}+2012\cdot 2k\pi\right)}\right]+[/dispmath][dispmath]+2^{1006}\left[\cos\left(\frac{2012\pi}{4}+2012\cdot 2k\pi\right)-\cancel{i\sin\left(\frac{2012\pi}{4}+2012\cdot 2k\pi\right)}\right]=[/dispmath][dispmath]=2^{1006}\cdot 2\cos\left(\frac{2012\pi}{4}+2012\cdot 2k\pi\right)=2^{1007}\cos\left(503\pi+2k\pi\right)=[/dispmath][dispmath]=2^{1007}\cos\left(\pi+2\cdot 251\pi+2k\pi\right)=2^{1007}\underbrace{\cos\left(\pi+2k\pi\right)}_{-1}=-2^{1007}[/dispmath]
čime smo dobili identično rešenje kao i na prethodni način.