Números complexos

Considerando o desenvolvimento $(1+i)^n$ calcule o valor da soma  $S=\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$ :
Observação : $i^2=-1$


Obtendo a expansão do binômio de Newton de $(1+i)^n$ , obtemos:
$$(1+i)^n$$=
$$\binom{n}{0}i^0+\binom{n}{1}i^1+\binom{n}{2}i^2+\binom{n}{3}i^3+\binom{n}{4}i^4+\binom{n}{5}i^5+\binom{n}{6}i^6+\cdots$$

Temos
$i^0=1$
$i^1=i$
$i^2=-1$
$i^3=-i$
$i^4=1$
$i^5=i$

Assim , substituindo na expansão, ficamos com:

$$\binom{n}{0}i+\binom{n}{1}i^1+\binom{n}{2}i^2+\binom{n}{3}i^3+\binom{n}{4}i^4+\binom{n}{5}i^5+\binom{n}{6}i^6+\cdots$$

Assim , temos:
$$\binom{n}{0}+\binom{n}{1}i -\binom{n}{2}-\binom{n}{3}i+\binom{n}{4}+\binom{n}{5}i-\binom{n}{6}+\cdots$$

Reorganizando:

$$\binom{n}{0}-\binom{n}{2} +i[\binom{n}{1}-\binom{n}{3}]+\binom{n}{4}-\binom{n}{6}-i[\binom{n}{5}-\binom{n}{7}]+\cdots$$

Colocando o número complexo  $(1+1)^n$ na forma trigonométrica , temos:
$$(1+i)^n=(\sqrt 2)^n (cos \frac{n\pi}{4}+i sen \frac{n\pi}{4})$$

Concluímos que $S=\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots$  é a parte real do número complexo $(1+1)^n$.
Logo
$$S=\binom{n}{0}-\binom{n}{2}+\binom{n}{4}-\binom{n}{6}+\cdots=(\sqrt 2)^n cos \frac{n\pi}{4}$$