Question

Calcule $(\frac{1}{2} -\frac{i}{2})^1^0^0$

Answer 1

$\mathrm{Seja}\ z=\dfrac{1}{2}-\dfrac{1}{2}i.\ \mathrm{Temos\ que}\ \Re{(z)}=a=\dfrac{1}{2}\ \text{e}\ \Im{(z)}=b=-\dfrac{1}{2}.$

$\mathrm{O\ m\acute{o}dulo\ de}\ z\ (\rho)\ \mathrm{ser\acute{a}}\ \text{dado por:}$

$\rho=\sqrt{a^2+b^2}=\sqrt{\bigg(\dfrac{1}{2}\bigg)^2+\bigg(-\dfrac{1}{2}\bigg)^2}=\sqrt{\dfrac{1}{2}}\ \therefore\ \boxed{\rho=\dfrac{1}{\sqrt{2}}}$

$\mathrm{Seja}\ \phi\ \mathrm{o\ argumento\ de}\ z\mathrm{,\ o\ cosseno\ e\ o\ seno\ de}\ \phi\ \mathrm{ser\tilde{a}}\text{o:}$

$\cos{\phi}=\dfrac{a}{\rho}=\dfrac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}\ \therefore\ \cos{\phi}=\dfrac{1}{\sqrt{2}}$

$\sin{\phi}=\dfrac{b}{\rho}=-\dfrac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}\ \therefore\ \sin{\phi}=-\dfrac{1}{\sqrt{2}}$

$\Longrightarrow \cos{\phi}=\dfrac{1}{\sqrt{2}}\ \text{e}\ \sin{\phi}=-\dfrac{1}{\sqrt{2}}\ \therefore\ \boxed{\phi=\dfrac{7\pi}{4}}$

$\mathrm{\acute{E}\ poss\acute{\i}vel\ escrever}\ z\ \text{em sua forma polar:}$

$z=\rho(\cos{\phi}+i\sin{\phi})\Longrightarrow \boxed{z=\dfrac{1}{\sqrt{2}}\bigg(\cos{\dfrac{7\pi}{4}}+i\sin{\dfrac{7\pi}{4}}\bigg)}$

$\mathrm{Para\ calcular}\ z^{100},\ \mathrm{\acute{e}\ preciso\ utilizar\ a\ F\acute{o}rmula\ de\ De}\ \text{Moivre:}$

$z^{n}=\rho^n(\cos{n\phi}+i\sin{n\phi})\Longrightarrow z^{100}=\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{100}\bigg(\cos{\dfrac{700\pi}{4}}+i\sin{\dfrac{700\pi}{4}}\bigg)$

$\Longrightarrow z^{100}=\dfrac{1}{2^{50}}(\cos{175\pi}+i\sin{175\pi})=\dfrac{1}{2^{50}}(\cos{\pi}+i\sin{\pi})$

$\Longrightarrow z^{100}=\dfrac{1}{2^{50}}(-1+i(0))\ \therefore\ \boxed{z^{100}=-\dfrac{1}{2^{50}}}$