Determine \arg _{0} z, onde z \in\left\{1+i,(1+i)^{4},\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^{5},\left(\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)^{5}\right\}. Calcule o logaritmo (ramo principal) desses números.

Passo 1

O enunciado pede para cada \;z\in\{\,1+i,\;(1+i)^4,\;(\tfrac{\sqrt{3}}{2}+i\tfrac{1}{2})^5,\;(\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2})^5\}
determinar o argumento principal (denotado por \Arg_0 z, com -\pi<\Arg_0 z\le\pi) e, em seguida, calcular o logaritmo principal \Log z = \ln|z| + i\,\Arg_0 z.
Vamos usar a forma polar \;z=r\,e^{i\theta}, a propriedade de De Moivre e ajustar \theta ao intervalo (-\pi,\pi].

Passo 2

Para z=1+i:
– Módulo:
|1+i|=\sqrt{1^2+1^2}=\sqrt{2}.
– Argumento: como \Re>0 e \Im>0, \Arg_0(1+i)=\tfrac{\pi}{4}.
– Logaritmo principal:
\Log(1+i)=\ln\bigl(\sqrt{2}\bigr)+i\,\tfrac{\pi}{4}.

Passo 3

Para z=(1+i)^4:
– Escrevemos 1+i=\sqrt{2}\,e^{i\pi/4}.
– Pela fórmula de De Moivre,
(1+i)^4=(\sqrt{2})^4\,e^{i4\cdot(\pi/4)} =4\,e^{i\pi} =-4.
– Aqui \Arg_0(-4)=\pi (ficando no limite superior do ramo).
– Logo
\Log\bigl((1+i)^4\bigr)=\ln(4)+i\,\pi.

Passo 4

Para z=\bigl(\tfrac{\sqrt{3}}{2}+i\tfrac{1}{2}\bigr)^5:
– Note que \tfrac{\sqrt{3}}{2}+i\tfrac{1}{2}=e^{i\pi/6}.
– Então
z=e^{i5\cdot(\pi/6)}=e^{i5\pi/6},
cujo argumento principal é \Arg_0(z)=5\pi/6 (já em (-\pi,\pi]) e |z|=1.
– Assim
\Log\bigl((\tfrac{\sqrt{3}}{2}+i\tfrac{1}{2})^5\bigr) =\ln(1)+i\,\tfrac{5\pi}{6} =i\,\tfrac{5\pi}{6}.

Passo 5

Para z=\bigl(\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2}\bigr)^5:
– Aqui \tfrac{1}{2}+i\tfrac{\sqrt{3}}{2}=e^{i\pi/3}.
– Então
z=e^{i5\cdot(\pi/3)}=e^{i5\pi/3}.
Mas 5\pi/3=300^\circ fica fora de (-\pi,\pi]; o ângulo equivalente é
5\pi/3 - 2\pi = -\pi/3,
logo \Arg_0(z)=-\pi/3 e |z|=1.
– Portanto
\Log\bigl((\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2})^5\bigr) =\ln(1)+i\,\bigl(-\tfrac{\pi}{3}\bigr) =-\,i\,\tfrac{\pi}{3}.

Resposta

Para cada z temos:
(a) z=1+i: \Arg_0(z)=\tfrac{\pi}{4}, \Log z=\ln\bigl(\sqrt{2}\bigr)+i\,\tfrac{\pi}{4}. (b) z=(1+i)^4: \Arg_0(z)=\pi, \Log z=\ln(4)+i\,\pi. (c) z=\bigl(\tfrac{\sqrt{3}}{2}+i\tfrac{1}{2}\bigr)^5: \Arg_0(z)=\tfrac{5\pi}{6}, \Log z=i\,\tfrac{5\pi}{6}. (d) z=\bigl(\tfrac{1}{2}+i\tfrac{\sqrt{3}}{2}\bigr)^5: \Arg_0(z)=-\tfrac{\pi}{3}, \Log z=-\,i\,\tfrac{\pi}{3}.

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