Verified answer

[tex]\boxed{\begin{matrix}\text{Propriedades de n\'umeros complexos}\\\\ \sf z = a+b\cdot i\\\\ \underline{\text{m\'odulo do complexo z}}:\\\\ \sf |z| = \sqrt{a^2+b^2} \\\\\ \sf z =|z|\cdot \underbrace{\sf \left[cos(\theta)+i\cdot sen(\theta)\right]}_{cis(\theta)} \to \sf z=|z|\cdot cis(\theta) \ ;\\\\\ \underline{\text{Potencia en\'esima de z}}: \\\\ \sf z^n =|z|^n\cdot cis(n\cdot \theta)\\\\ \sf onde: \\\\ \displaystyle \sf cos(\theta)=\frac{a}{|z|}\ ;\ sen(\theta) =\frac{b}{|z|}\ ; \\_\end{matrix}}[/tex]

Queremos mostrar que para n > 1 é válido :

[tex]\displaystyle \sf 1-{n\choose 2}+{n\choose 4}-{n\choose 6}+... = 2^{\frac{n}{2}}\cdot cos\left(\frac{n\cdot \pi}{4}\right)[/tex]

Sabemos que :

[tex]\displaystyle \sf (1+x)^{n} = \sum^{n}_{p=0} {n\choose p} \cdot 1^{n-p}\cdot x^{p} \\\\\\ (1+x)^{n} = {n\choose 0} \cdot x^{0}+{n\choose 1}\cdot x^{1}+{n\choose 2}\cdot x^{2}+... {n\choose n} \cdot x^{n} \\\\\ \text{Fa\c camos} : \\\\ x = i \to \text{(unidade complexa)} \\\\\ (1+i)^{n} = {n\choose 0}\cdot 1+{n\choose 1}\cdot i+{n\choose 2}\cdot i^2+{n\choose 3}\cdot i^3 +{n\choose 4}\cdot i^4+...+{n\choose n}\cdot i^{n} \\\\\\\[/tex]

[tex]\displaystyle \sf \boxed{\begin{matrix}\text{Obs:} \\\\ \sf z = 1+i\\\\ \sf |z| = \sqrt{1^2+1^2} =\sqrt{2} \\\\\ \displaystyle \sf z = \sqrt{2}\cdot cis\left(\frac{\pi}{4}\right) \end{matrix}} \\\\\\ \sf Da{\'i}}: \\\\\ z^{n}=\sqrt{2}^{n}\cdot cis\left(\frac{n\cdot \pi}{4}\right) \\\\\\\[/tex]

[tex]\displaystyle \sf z^{n} = \underbrace{\sf 2^{\frac{n}{2}}\cdot cos\left(\frac{n\cdot \pi}{4}\right)}_{\displaystyle \text{Parte real}}+\underbrace{\sf 2^{\frac{n}{2}}\cdot i\cdot sen\left(\frac{n\cdot \pi}{4}\right)}_{\displaystyle \text{Parte imagin\'aria}}[/tex]

Daí, temos :

[tex]\displaystyle \sf (1+i)^{n} = 1+{n\choose 1}\cdot i+{n\choose 2}\cdot (-1)+{n\choose 3}\cdot (-i) +{n\choose 4}\cdot 1+...+{n\choose n}\cdot i^{n} \\\\\\ 2^{\frac{n}{2}}\cdot cis\left(\frac{n\cdot \pi}{4}\right) = 1+{n\choose 1}\cdot i-{n\choose 2}-{n\choose 3}\cdot i +{n\choose 4}+...+{n\choose n}\cdot i^{n} \\\\\\[/tex]

[tex]\displaystyle \sf 2^{\frac{n}{2}}\cdot cis\left(\frac{n\cdot \pi}{4}\right) = \underbrace{\sf \left[1-{n\choose 2}+{n\choose 4}-...\right]}_{\displaystyle \text{Parte real}} + \underbrace{\sf i\cdot \left[{n\choose 1}-{n\choose 3}+{n\choose 5}-...\right]}_{\displaystyle \text{Parte imagin\'aria}}[/tex]

Igualando parte real com parte real :

[tex]\displaystyle \sf \boxed{\sf \ 2^{\frac{\sf n}{2}}\cdot cos\left(\frac{n\cdot \pi}{4}\right) = 1-{n\choose 2}+{n\choose 4}-{n\choose 6}+..\ }\checkmark[/tex]