Determine os logaritmos dos numerous seguintes: A) e; B) -i; C) i^i; D) 1-i.... Pergunta de ideia deRonny06

January 2020 1 145 Report

Determine os logaritmos dos numerous seguintes:

A) e;

B) -i;

C) i^i;

D) 1-i

Lista de comentários

Lukyo

Verified answer

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_______________

Dado um número complexo $\mathsf{z},$ sabemos que

$\mathsf{z=Re(z)+i\cdot Im(z)}$

sendo $\mathsf{Re(z)}$ e $\mathsf{Im(z)}$ respectivamente as partes real e imaginária de $\mathsf{z.}$

Podemos expressar $\mathsf{z}$ na forma polar (ou trigonométrica):

$\mathsf{z=|z|\cdot e^{i\cdot \theta}=|z|\cdot (cos\,\theta + i\cdot sen\,\theta)}$

sendo $\mathsf{|z|}$ o módulo de $\mathsf{z},$ dado por

$\mathsf{|z|=\sqrt{Re^2(z)+Im^2(z)}}$

e $\mathsf{\theta=Arg(z)}$ o argumento de $\mathsf{z}.$ Este ângulo $\theta$ é tal que

$\left\{\!
\begin{array}{l}
\mathsf{cos\,\theta=\dfrac{Re(z)}{|z|}=\dfrac{Re(z)}{\sqrt{Re^2(z)+Im^2(z)}}}\\\\\\
\mathsf{sen\,\theta=\dfrac{Im(z)}{|z|}=\dfrac{Im(z)}{\sqrt{Re^2(z)+Im^2(z)}}}
\end{array}
\right.\qquad\qquad\mathsf{-\pi\ \textless \ \theta\le \pi}$

Nessas condições, o logaritmo de $\mathsf{z}$ é dado por

$\boxed{\begin{array}{c}\mathsf{\ell n(z)=\ell n|z| + i\cdot \theta}\end{array}}$

__________

A) $\mathsf{z=e}$

Aqui temos um número real. Como $\mathsf{z=e}$ é positivo, o argumento é igual a zero (se fosse negativo, o argumento seria $\mathsf{\pi}$ ).

O valor do logaritmo é o mesmo obtido enquanto trabalhamos com reais:

$\mathsf{\ell n(e)=\ell n|e| + i\cdot 0}\\\\
\mathsf{\ell n(e)=1 + 0i}\\\\
\mathsf{\ell n(e)=1\qquad\quad\checkmark}$

________

B) $\mathsf{z=-i}$

$\mathsf{z=0-1i}$

Calculando o módulo e o argumento:

$\mathsf{|z|=\sqrt{0^2+(-1)^2}}\\\\
\mathsf{|z|=\sqrt{0+1}}\\\\
\mathsf{|z|=\sqrt{1}}\\\\
\mathsf{|z|=1}$

$\left\{\! \begin{array}{l} \mathsf{cos\,\theta=\dfrac{0}{1}=0}\\\\\\ \mathsf{sen\,\theta=\dfrac{-1}{1}=-1} \end{array} \right.\qquad\Rightarrow\qquad\mathsf{\theta=-\dfrac{\pi}{2}}$

Portanto,

$\mathsf{\ell n(-i)=\ell n(1)+i\cdot \left(-\dfrac{\pi}{2}\right)}\\\\\\
\mathsf{\ell n(-i)=0-i\cdot \dfrac{\pi}{2}}\\\\\\
\mathsf{\ell n(-i)=-\dfrac{\pi}{2}\,i}\\\\\\ \mathsf{\ell n(-i)\approx -1,\!571\,i\qquad\quad\checkmark}$

________

C) $\mathsf{z=i^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i}\end{array}}}$

$\mathsf{z=(1+0i)^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i}\end{array}}}\\\\
\mathsf{z=\left(cos\,\dfrac{\pi}{2}+i\cdot sen\,\dfrac{\pi}{2}\right)^{\hspace{-6}\footnotesize\begin{array}{l}\mathsf{i}\end{array}}}\\\\\\
\mathsf{z=\left(e^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i\cdot \frac{\pi}{2}}\end{array}\hspace{-5}}\right)^{\hspace{-6}\footnotesize\begin{array}{l}\mathsf{i}\end{array}}}$

$\mathsf{z=e^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i^2\cdot \frac{\pi}{2}}\end{array}\hspace{-5}}}\\\\
\mathsf{z=e^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{(-1)\cdot \frac{\pi}{2}}\end{array}\hspace{-5}}}\\\\
\mathsf{z=e^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{-\frac{\pi}{2}}\end{array}\hspace{-5}}\in\mathbb{R}}$

Aqui temos um número real. Como $\mathsf{z=e^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{-\frac{\pi}{2}}\end{array}\hspace{-5}}}$ é positivo, o argumento é igual a zero.

Portanto,

$\mathsf{\ell n\big(i^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i}\end{array}\hspace{-5}}\big)=\ell n\big(e^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{-\frac{\pi}{2}}\end{array}\hspace{-5}}\big)}\\\\\\
\mathsf{\ell n\big(i^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i}\end{array}\hspace{-5}}\big)=-\dfrac{\pi}{2}\cdot \ell n(e)}\\\\\\
\mathsf{\ell n\big(i^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i}\end{array}\hspace{-5}}\big)=-\dfrac{\pi}{2}\cdot 1}$

$\mathsf{\ell n\big(i^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i}\end{array}\hspace{-5}}\big)=-\dfrac{\pi}{2}}\\\\\\ \mathsf{\ell n\big(i^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{i}\end{array}\hspace{-5}}\big)\approx -1,\!571}\qquad\quad\checkmark}$

________

D) $\mathsf{z=1-i}$

Calculando o módulo e o argumento:

$\mathsf{|z|=\sqrt{1^2+(-1)^2}}\\\\ \mathsf{|z|=\sqrt{1+1}}\\\\ \mathsf{|z|=\sqrt{2}}}$

$\left\{\! \begin{array}{l} \mathsf{cos\,\theta=\dfrac{1}{\sqrt{2}}}\\\\\\ \mathsf{sen\,\theta=\dfrac{-1}{\sqrt{2}}} \end{array} \right.\qquad\Rightarrow\qquad\mathsf{\theta=-\dfrac{\pi}{4}}$

Portanto,

$\mathsf{\ell n(1-i)=\ell n\big(\sqrt{2}\big)+i\cdot \left(-\dfrac{\pi}{4}\right)}\\\\\\
\mathsf{\ell n(1-i)=\ell n\big(2^{\hspace{-4}\footnotesize\begin{array}{l}\mathsf{\frac{1}{2}}\end{array}\hspace{-5}}\big)-i\cdot \dfrac{\pi}{4}}\\\\\\
\mathsf{\ell n(1-i)=\dfrac{1}{2}\cdot\ell n(2)-i\cdot \dfrac{\pi}{4}}\\\\\\
\mathsf{\ell n(1-i)\approx 0,\!347-0,\!785\,i\qquad\quad\checkmark}$

Bons estudos!:-)

Tags: logaritmo real imaginária módulo argumento ângulo número complexo

2 votes Thanks 2

Ronny06 Matematico, -I= (-1)*I logo nao conclui-se que -I= e^3*pi/2?

Lukyo O argumento deve estar no intervalo de -pi a pi.

Lukyo Sim, a igualdade que você escreveu é verdadeira, mas cada número complexo só pode ter um logaritmo.

Lukyo -i = e^3pi/2 = e^-pi/2.

Lukyo perdão, esqueci do i no expoente.

Lukyo -i = e^(i * 3pi/2) = e^(i * -pi/2).

Ronny06 Percebido !! Sempre esqueco da condicao do argumento, que em algum momento sempre tenho que acabar por reduzir neste caso (270-180=90) Agora so nao entende o porque do sinal " - "

Ronny06 porque -90 ao inves de 90?

Lukyo 270 - 360 = -90

Lukyo vc tem que reduzir somando ou subtraindo múltiplos de 360 graus