Question

06. A potência (1 - i)^16 equivale a:
resposta com cálculo.

a) 8
b) 16 - 4i
c) 16 - 16i
d) 256 - 16i
e) 256​

Answer 1

$\Large\boxed{\begin{array}{l}\rm Z=a+bi\\\underline{\sf M\acute odulo~de~um~n\acute umero~complexo}\\\huge\boxed{\boxed{\boxed{\boxed{\rm\rho=\sqrt{a^2+b^2} }}}}\\\underline{\sf Argumento~de~um~n\acute umero~complexo}\\\rm \acute E~o~\hat angulo~\theta~tal~que\\\rm sen(\theta)=\dfrac{a}{\rho}~e~cos(\theta)=\dfrac{b}{\rho}\\\underline{\sf Forma~trigonom\acute etrica~de~um~n\acute umero~complexo}\\\huge\boxed{\boxed{\boxed{\boxed{\rm Z=\rho[cos(\theta)+i~sen(\theta)]}}}}\end{array}}$

$\large\boxed{\begin{array}{l}\rm z=1-i\\\underline{\sf C\acute alculo\,do\,m\acute odulo}\\\rm \rho=\sqrt{1^2+(-1)^2}\\\rm \rho=\sqrt{1+1}\\\rm\rho=\sqrt{2}\\\underline{\sf C\acute alculo\,do\,argumento}\\\begin{cases}\rm cos(\theta)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\rm sen(\theta)=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\end{cases}\longrightarrow \theta=\dfrac{7\pi}{4}\\\\\rm \end{array}}$

$\large\boxed{\begin{array}{l}\rm(1-i)^{16}=z^{16}=\rho^{16}[cos(16\theta)+i~sen(16\theta)]\\\\\rm (1-i)^{16}=\sqrt{2^{16}}\cdot\bigg[cos\bigg(16\cdot\dfrac{7\pi}{4}\bigg)+i~sen\bigg(16\cdot\dfrac{7\pi}{4}\bigg)\\\\\rm (1-i)^{16}=2^8[cos(28\pi)+i~sen(28\pi)] \\\rm 28|\underline{2~~~~}~\\\rm\!\!\!\!-\underline{28}~~14\\\rm~0\\\rm cos(28\pi)=cos(0)=1\\\rm (1-i)^{16}=256[1+i\cdot0]\\\\\rm (1-i)^{16}=256\\\huge\boxed{\boxed{\boxed{\boxed{\rm\dagger\red{\maltese}~\blue{alternativa~e}}}}}\end{array}}$