Question

Considere o número complexo Z = 1 + √3i. Determine:
a) O módulo de Z
b) O argumento de Z
c) Z⁶
d) As raízes quadradas de Z

Apresente a resolução:

Answer 1

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Módulo de um número complexo

$\Large\boxed{\begin{array}{l}\sf seja~z=a+bi~um~n\acute umero~complexo.\\\sf o~m\acute odulo~de~z~\acute e\\\huge\boxed{\boxed{\boxed{\boxed{\sf \rho=\sqrt{a^2+b^2}}}}}\end{array}}$

Argumento de um número complexo

$\Large\boxed{\begin{array}{l}\sf\acute e~o~\hat angulo~\theta~tal~que\\\sf cos(\theta)=\dfrac{a}{\rho}~e~sen(\theta)=\dfrac{b}{\rho}\end{array}}$

Potenciação de números complexos

$\huge\boxed{\boxed{\boxed{\boxed{\sf z^n=\rho^n[cos(n\theta)+i~sen(n\theta)]}}}}$

Raízes enésimas de um número complexo

$\large\boxed{\boxed{\boxed{\boxed{\sf w_k=\sqrt[\sf n]{\sf\rho}\bigg[cos\bigg(\dfrac{\theta+2k\pi}{n}\bigg)+i~sen\bigg(\dfrac{\theta+2k\pi}{n}\bigg)\bigg]}}}}$

$\sf z=1+\sqrt{3}i\\\tt a)~\sf \rho=\sqrt{1^2+(\sqrt{3})^2}\\\sf \rho=\sqrt{1+3}\\\sf \rho=\sqrt{4}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\rho=2}}}}\blue{\checkmark}\\\tt b)~\sf tg(\theta)=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\implies \theta=arctg\bigg(\dfrac{\sqrt{3}}{3}\bigg)\\\huge\boxed{\boxed{\boxed{\boxed{\sf\theta=\dfrac{\pi}{6}}}}}\blue{\checkmark}$

$\tt c)~\sf z^6=2^6\bigg[cos\bigg(\backslash\!\!\!6\cdot\dfrac{\pi}{\backslash\!\!\!6}\bigg)+i~sen\bigg(\backslash\!\!\!6\cdot\dfrac{\pi}{\backslash\!\!\!6}\bigg)\bigg]\\\sf z^6=64[cos(\pi)+i~sen(\pi)]\\\sf z^6=64[-1]\\\huge\boxed{\boxed{\boxed{\boxed{\sf z=-64}}}}\blue{\checkmark}$

$\tt d)~\sf w_k=\sqrt{2}\bigg[cos\bigg(\dfrac{\frac{\pi}{6}+2k\pi}{2}\bigg)+i~sen\bigg(\dfrac{\frac{\pi}{6}+2k\pi}{2}\bigg)\bigg]\\\sf w_k=\sqrt{2}\bigg[cos\bigg(\dfrac{\pi}{12}+k\pi\bigg)+i~sen\bigg(\dfrac{\pi}{12}+k\pi\bigg)\bigg]\\\sf se~k=0:\\\sf w_0=\sqrt{2}\bigg[cos\bigg(\dfrac{\pi}{12}\bigg)+i~sen\bigg(\dfrac{\pi}{12}\bigg)\bigg]\\\sf se~k=1:\\\sf w_1=\sqrt{2}\bigg[cos\bigg(\dfrac{\pi}{12}+\pi\bigg)+i~sen\bigg(\dfrac{\pi}{12}+\pi\bigg)\bigg]$

$\sf w_1=\sqrt{2}\bigg[cos\bigg(\dfrac{13\pi}{12}\bigg)+i~sen\bigg(\dfrac{13\pi}{12}\bigg)\bigg]$

$\sf k=2\\\sf w_2=\sqrt{2}\bigg[cos\bigg(\dfrac{\pi}{12}+2\pi\bigg)+i~sen\bigg(\dfrac{\pi}{12}+2\pi\bigg)\bigg]\\\sf w_2=\sqrt{2}\bigg[cos\bigg(\dfrac{25\pi}{12}\bigg)+i~sen\bigg(\dfrac{25\pi}{12}\bigg)\bigg]$

$\boxed{\begin{array}{l}\sf portanto~as~ra\acute izes~quadradas~de~Z=1+\sqrt{3}i\\\sf s\tilde ao:\\\huge\boxed{\boxed{\boxed{\boxed{\sf\sqrt{2}\bigg[cos\bigg(\dfrac{\pi}{12}\bigg)+i~sen\bigg(\dfrac{\pi}{12}\bigg)\bigg]}}}}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\sqrt{2}\bigg[cos\bigg(\dfrac{13\pi}{12}\bigg)+i~sen\bigg(\dfrac{13\pi}{12}\bigg)\bigg]}}}}\\\rm e\end{array}}$

$\boxed{\begin{array}{l}\huge\boxed{\boxed{\boxed{\boxed{\sf\sqrt{2}\bigg[cos\bigg(\dfrac{25\pi}{12}\bigg)+i~sen\bigg(\dfrac{25\pi}{12}\bigg)\bigg]}}}}\end{array}}$