Question

qual o valor de (1-i)⁶⁸?​

Answer 1

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

$\large\boxed{\begin{array}{l}\rm z=1-i\\\rm \rho=\sqrt{1^2+(-1)^2}\\\rm \rho=\sqrt{1+1}\\\rm \rho=\sqrt{2}\\\begin{cases}\rm cos(\theta)=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\\\\rm sen(\theta)=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\end{cases}\longrightarrow \theta=\dfrac{7\pi}{4}\\\\\rm (1-i)^{68}=z^{68}\\\sf z^{68}=\rho^{68}\cdot[cos(68\theta)+i~sen(68\theta)]\\\rm z^{68}=(\sqrt{2})^{68}\cdot\bigg[cos\bigg(68\cdot\dfrac{7\pi}{4}\bigg)+i~sen\bigg(68\cdot\dfrac{7\pi}{4}\bigg)\bigg]\end{array}}$

$\large\boxed{\begin{array}{l}\rm z^{68}=2^{34}\cdot(-1+i~\cdot0)\\\rm z^{68}=-2^{34}\end{array}}$