Ensino Superior ⇒ Trigonometria Imagens das Funções.

[eliz2016]

Boa tarde peço ajuda nessa questão por favor.

Considere os conjuntos A=[tex3{z}\in \mathbb {C}\mathbb /{R}e(z) [/tex3] = [tex3]Im(z)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]Im(z)[/tex3]

e B={z [tex3]\in \mathbb{C}/Im(z)\geq 0 [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\in \mathbb{C}/Im(z)\geq 0 [/tex3]

[tex3]e \left | z \right |=1[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]e \left | z \right |=1[/tex3]

.
Determine as imagens de A e B pelas funções:

a)[tex3]f(z)=\bar{z}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]f(z)=\bar{z}[/tex3]

[rcompany]

Boa tarde peço ajuda nessa questão por favor.

Considere os conjuntos A=[tex3]\{z\in \mathbb{C}/\mathrm{Re}(z) =\mathrm{Im}(z)[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\{z\in \mathbb{C}/\mathrm{Re}(z) =\mathrm{Im}(z)[/tex3]

e [tex3]B=\{z \in \mathbb{C}/\mathrm{Im}(z)\geq 0\text{ e }\left | z \right |=1\}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]B=\{z \in \mathbb{C}/\mathrm{Im}(z)\geq 0\text{ e }\left | z \right |=1\}[/tex3]

.
Determine as imagens de A e B pelas funções:

a)[tex3]f(z)=\bar{z}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]f(z)=\bar{z}[/tex3]

[tex3]\begin{array}{rl}\forall z \in\mathbb{C},z\in A&\iff z=a+ia,\ a\in\mathbb{R}\\
&\iff f(z)=|z|=\sqrt{2a^2}=|a|\sqrt{2}\\
&\implies f(z)\in\mathbb{R}^+\end{array}\\[24pt]

\begin{array}{rl}
\forall x in \mathbb{R}^+,x&=\sqrt{x^2}=\sqrt{\dfrac{1}{2}x^2+\dfrac{1}{2}x^2}\\&=\sqrt{\left(\dfrac{x}{\sqrt{2}}\right)^2+\left(\dfrac{x}{\sqrt{2}}\right)^2}\\
&=\left|\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}}\right|\implies (\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}})\in A
\end{array}\\[48pt]
\text{ e então }\forall x\in\mathbb{R}^+, \exists z\in \!A/f(z)=x\\[12pt]
\text{Conclusão: }f(A)=\mathbb{R}^+\\[24pt]
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\begin{array}{rl}\forall z \in\mathbb{C},z\in A&\iff z=a+ia,\ a\in\mathbb{R}\\
&\iff f(z)=|z|=\sqrt{2a^2}=|a|\sqrt{2}\\
&\implies f(z)\in\mathbb{R}^+\end{array}\\[24pt]

\begin{array}{rl}
\forall x in \mathbb{R}^+,x&=\sqrt{x^2}=\sqrt{\dfrac{1}{2}x^2+\dfrac{1}{2}x^2}\\&=\sqrt{\left(\dfrac{x}{\sqrt{2}}\right)^2+\left(\dfrac{x}{\sqrt{2}}\right)^2}\\
&=\left|\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}}\right|\implies (\dfrac{x}{\sqrt{2}}+i\dfrac{x}{\sqrt{2}})\in A
\end{array}\\[48pt]
\text{ e então }\forall x\in\mathbb{R}^+, \exists z\in \!A/f(z)=x\\[12pt]
\text{Conclusão: }f(A)=\mathbb{R}^+\\[24pt]
[/tex3]

[tex3]
\begin{array}{rl}

\forall z\in \mathbb{C}, z\in B &\iff\left\{\begin{array}{l}z=a+ib,\ a\in\mathbb{R},b\in\mathbb{R}^+\\a^2+b^2=1\end{array}\right.\\
&\implies f(z)=1\\
\text{e então }f(B)=\{1\}
\end{array}\\[36pt]

\text{Se o enunciado for }B=\{z\in\mathbb{C}/\mathrm{Im}(z)\geqslant 0\}:\\[24pt]

\begin{array}{rl}

\forall z\in \mathbb{C}, z\in B &\iff\left\{\begin{array}{l}z=a+ib,\ a\in\mathbb{R},b\in\mathbb{R}^+\\b\geqslant 0\end{array}\right.\\
&\implies f(z)=\sqrt{a^2+b^2}\\
&\implies f(z)\in\mathbb{R}^+
\end{array}\\[36pt]
\begin{array}{rl}
\forall x\in\mathbb{R}^+, \forall b\in[0;x], x&=\sqrt{x^2}\\&=\sqrt{x^2-b^2+b^2}\\&=\sqrt{\left(\sqrt{x^2-b^2}\right)^2+b^2}\\
&=\left|\sqrt{x^2-b^2}+ib\right|\quad\text{e }b\geqslant 0\\

\end{array}\\
\text{E então }\forall x \in\mathbb{R}^+,\exists z\in\ \!\!B/f(z)=x\\[24pt]
\text{Conclusão:}f(B)=\mathbb{R}^+
[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]
\begin{array}{rl}

\forall z\in \mathbb{C}, z\in B &\iff\left\{\begin{array}{l}z=a+ib,\ a\in\mathbb{R},b\in\mathbb{R}^+\\a^2+b^2=1\end{array}\right.\\
&\implies f(z)=1\\
\text{e então }f(B)=\{1\}
\end{array}\\[36pt]

\text{Se o enunciado for }B=\{z\in\mathbb{C}/\mathrm{Im}(z)\geqslant 0\}:\\[24pt]

\begin{array}{rl}

\forall z\in \mathbb{C}, z\in B &\iff\left\{\begin{array}{l}z=a+ib,\ a\in\mathbb{R},b\in\mathbb{R}^+\\b\geqslant 0\end{array}\right.\\
&\implies f(z)=\sqrt{a^2+b^2}\\
&\implies f(z)\in\mathbb{R}^+
\end{array}\\[36pt]
\begin{array}{rl}
\forall x\in\mathbb{R}^+, \forall b\in[0;x], x&=\sqrt{x^2}\\&=\sqrt{x^2-b^2+b^2}\\&=\sqrt{\left(\sqrt{x^2-b^2}\right)^2+b^2}\\
&=\left|\sqrt{x^2-b^2}+ib\right|\quad\text{e }b\geqslant 0\\

\end{array}\\
\text{E então }\forall x \in\mathbb{R}^+,\exists z\in\ \!\!B/f(z)=x\\[24pt]
\text{Conclusão:}f(B)=\mathbb{R}^+
[/tex3]

[eliz2016]

Boa noite muito obrigada.