expansão binomial - Politécnitca

Pelo binômio de newton:

[latex](1+i)^n = \binom{n}{0}i^0+ \binom{n}{1}i^1+ \binom{n}{2}i^2+ \binom{n}{3}i^3+...[/latex]

[latex](1+i)^n = \binom{n}{0}+ \binom{n}{1}i- \binom{n}{2}- \binom{n}{3}i+...[/latex]

[latex](1+i)^n = (\binom{n}{0}- \binom{n}{2} +\binom{n}{4}-...) +i(\binom{n}{1}-\binom{n}{3}+\binom{n}{5}...)[/latex]

[latex](1+i)^n = a+bi[/latex]

Ora, mas também podemos escrever:

[latex](1+i)^n = (\sqrt{2}cis\frac{\pi }{4})^n=2^{\frac{n}{2}}*cis\frac{n\pi }{4}[/latex]

[latex](1+i)^n = 2^{\frac{n}{2}}*cos\frac{n\pi }{4}+i*2^{\frac{n}{2}}*sen\frac{n\pi }{4}[/latex]

Comparando as expressões, vem

[latex]a=2^{\frac{n}{2}}*cos\frac{n\pi }{4}\: \: e\: \: \: b=2^{\frac{n}{2}}*sen\frac{n\pi }{4}[/latex]

De modo que [latex]a^2+b^2 = 2^n[/latex].