Qual é a resposta correta?

Seja $z = {\cos ^2}\theta  + \frac{i}{2}\operatorname{sen} \left( {2\theta } \right)$ e $\theta  \in \left] { – \pi ,\pi } \right[$.

Exercício 1

O módulo de $z$ é:

[A] $\cos \theta $, qualquer que seja $\theta $;

[B] $\cos \theta $, se $\theta  \in \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]$;

[C] $\cos \left( {\theta  + \pi } \right)$, se $\theta  \notin \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]$.

Ora, $$\begin{array}{*{20}{l}}
{\left| {\text{z}} \right|}& = &{\sqrt {{{\left( {{{\cos }^2}\theta } \right)}^2} + {{\left( {\frac{1}{2}\operatorname{sen} \left( {2\theta } \right)} \right)}^2}} } \\
{}& = &{\sqrt {{{\cos }^4}\theta  + \frac{1}{4}{{\operatorname{sen} }^2}\left( {2\theta } \right)} } \\
{}& = &{\sqrt {{{\cos }^4}\theta  + \frac{1}{4}{{\left( {\operatorname{sen} \left( {2\theta } \right)} \right)}^2}} } \\
{}& = &{\sqrt {{{\cos }^4}\theta  + \frac{1}{4}{{\left( {2\operatorname{sen} \theta \cos \theta } \right)}^2}} } \\
{}& = &{\sqrt {{{\cos }^4}\theta  + \frac{1}{4} \times 4{{\operatorname{sen} }^2}\theta  \times {{\cos }^2}\theta } } \\
{}& = &{\sqrt {{{\cos }^2}\theta \left( {{{\cos }^2}\theta  + {{\operatorname{sen} }^2}\theta } \right)} } \\
{}& = &{\sqrt {{{\cos }^2}\theta } } \\
{}& = &{\left| {\cos \theta } \right|} \\
{}& = &{\left\{ {\begin{array}{*{20}{l}}
{\cos \theta }& \Leftarrow &{\theta  \in \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]} \\
{\cos \left( {\pi  + \theta } \right)}& \Leftarrow &{\theta  \in \left( {\left] { – \pi ,\pi } \right[\backslash \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]} \right)}
\end{array}} \right.}
\end{array}$$

Logo, são corretas as respostas B e C.

Exercício 2

Um argumento de $z$ é:

[A] $\theta $, qualquer que seja $\theta $;

[B] $2\theta $:

[C] $\theta  + \pi $, se $\theta  \notin \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]$.

Ora, $$\begin{array}{*{20}{l}}
z& = &{{{\cos }^2}\theta  + \frac{i}{2}\operatorname{sen} \left( {2\theta } \right)} \\
{}& = &{{{\cos }^2}\theta  + \frac{i}{2} \times 2\operatorname{sen} \theta \cos \theta } \\
{}& = &{\cos \theta \left( {\cos \theta  + i\operatorname{sen} \theta } \right)}
\end{array}$$

Como ${\cos ^2}\theta  \geqslant 0,\forall \left] { – \pi ,\pi } \right[$, então $\arg z \in \left[ { – \frac{\pi }{2} + 2k\pi ,\frac{\pi }{2} + 2k\pi } \right],k \in \mathbb{Z}$

Assim:

• um argumento de $z$ é $\theta $ se ${\theta  \in \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]}$.
• um argumento de $z$ é $\theta  + \pi $ se ${\theta  \in \left( {\left] { – \pi ,\pi } \right[\backslash \left[ { – \frac{\pi }{2},\frac{\pi }{2}} \right]} \right)}$

Logo, apenas a resposta C é correta.

Exercício 3

${z^2}$ é igual a:

[A] ${\cos ^4}\theta  – \frac{1}{4}{\operatorname{sen} ^2}2\theta  + i\operatorname{sen} 2\theta {\cos ^2}\theta $;

[B] ${\cos ^2}\theta \operatorname{cis} \left( {2\theta } \right)$;

[C] ${\cos ^2}\left( {\theta  + \pi } \right)\operatorname{cis} \left[ {2\left( {\theta  + \pi } \right)} \right]$.

$$\begin{array}{*{20}{l}}
{{z^2}}& = &{{{\left( {{{\cos }^2}\theta  + \frac{i}{2}\operatorname{sen} \left( {2\theta } \right)} \right)}^2}}&{}&{} \\
{}& = &{{{\cos }^4}\theta  – \frac{1}{4}{{\operatorname{sen} }^2}\left( {2\theta } \right) + i\operatorname{sen} \left( {2\theta } \right){{\cos }^2}\theta }&{}&{{\text{[A]}}} \\
{}& = &{{{\cos }^4}\theta  – \frac{1}{4}{{\left( {2\operatorname{sen} \theta \cos \theta } \right)}^{\text{2}}} + i\operatorname{sen} \left( {2\theta } \right){{\cos }^2}\theta }&{}&{} \\
{}& = &{{{\cos }^2}\theta \left( {{{\cos }^2}\theta  – {{\operatorname{sen} }^2}\theta } \right) + i\operatorname{sen} \left( {2\theta } \right){{\cos }^2}\theta }&{}&{} \\
{}& = &{{{\cos }^2}\theta \cos \left( {2\theta } \right) + i\operatorname{sen} \left( {2\theta } \right){{\cos }^2}\theta }&{}&{} \\
{}& = &{{{\cos }^2}\theta \left( {\cos \left( {2\theta } \right) + i\operatorname{sen} \left( {2\theta } \right)} \right)}&{}&{} \\
{}& = &{{{\cos }^2}\theta \operatorname{cis} \left( {2\theta } \right)}&{}&{{\text{[B]}}} \\
{}& = &{{{\cos }^2}\left( {\theta  + \pi } \right)\operatorname{cis} \left( {2\theta  + 2\pi } \right)}&{}&{} \\
{}& = &{{{\cos }^2}\left( {\theta  + \pi } \right)\operatorname{cis} \left( {2\left( {\theta  + \pi } \right)} \right)}&{}&{{\text{[C]}}}
\end{array}$$

Logo, todas as respostas são corretas.