Question

Calcule o modulo o argumento e de a representaçao grafica dos numeros complexos:
a) z=1+√3i
b) z= -5 - 5i
c) z=-4√3 - 4i
d) z= 1-i
e)z=4i
f)z=2+2√3i

Answer 1

Para um número complexo $z=a+bi$, onde $a,b \in \mathbb{R}$, temos que

o módulo de $z$ é

$\left|z\right|=\sqrt{a^{2}+b^{2}}$


o argumento de $z$ é o ângulo $\theta_{z}$,  onde $0 \leq \theta_{z} <2\pi$,  tal que

$\left\{ \begin{array}{c} \cos \theta_{z}=\dfrac{a}{\left|z\right|}=\dfrac{a}{\sqrt{a^{2}+b^{2}}}\\ \\ \mathrm{sen\,}\theta_{z}=\dfrac{b}{\left|z\right|}=\dfrac{b}{\sqrt{a^{2}+b^{2}}} \end{array} \right.$


a) $z=1+\sqrt{3}i$

$a=1\\ \\ b=\sqrt{3}\\ \\ \\ \left|z\right|=\sqrt{\left(1\right)^{2}+\left(\sqrt{3}\right)^{2}}\\ \\ \left|z\right|=\sqrt{1+3}\\ \\ \left|z\right|=\sqrt{4}\\ \\ \boxed{\left|z\right|=2}\\ \\ \\ \left\{ \begin{array}{l} \cos \theta_{z}=\dfrac{1}{2}\\ \\ \mathrm{sen\,}\theta_{z}=\dfrac{\sqrt{3}}{2} \end{array}\right. \Rightarrow \boxed{\theta_{z}=\dfrac{\pi}{6}}$


b) $z=-5-5i$

$a=-5\\ \\ b=-5\\ \\ \\ \left|z\right|=\sqrt{\left(-5\right)^{2}+\left(-5\right)^{2}}\\ \\ \left|z\right|=\sqrt{25+25}\\ \\ \left|z\right|=\sqrt{25 \times 2}\\ \\ \boxed{\left|z\right|=5\sqrt{2}}\\ \\ \\ \left\{ \begin{array}{l} \cos \theta_{z}=\dfrac{-5}{5\sqrt{2}}=-\dfrac{1}{\sqrt{2}}\\ \\ \mathrm{sen\,}\theta_{z}=\dfrac{-5}{5\sqrt{2}}=-\dfrac{1}{\sqrt{2}} \end{array}\right. \Rightarrow \boxed{\theta_{z}=\dfrac{5\pi}{4}}$


c) $z=-4\sqrt{3}-4i$

$a=-4\sqrt{3}\\ \\ b=-4\\ \\ \\ \left|z\right|=\sqrt{\left(-4\sqrt{3}\right)^{2}+\left(-4\right)^{2}}\\ \\ \left|z\right|=\sqrt{48+16}\\ \\ \left|z\right|=\sqrt{64}\\ \\ \boxed{\left|z\right|=8}\\ \\ \\ \left\{ \begin{array}{l} \cos \theta_{z}=\dfrac{-4\sqrt{3}}{8}=-\dfrac{\sqrt{3}}{2}\\ \\ \mathrm{sen\,}\theta_{z}=\dfrac{-4}{8}=-\dfrac{1}{2} \end{array}\right. \Rightarrow \boxed{\theta_{z}=\dfrac{4\pi}{3}}$


d) $z=1-i$

$a=1\\ \\ b=-1\\ \\ \\ \left|z\right|=\sqrt{\left(1\right)^{2}+\left(-1\right)^{2}}\\ \\ \left|z\right|=\sqrt{1+1}\\ \\ \boxed{\left|z\right|=\sqrt{2}}\\ \\ \\ \left\{ \begin{array}{l} \cos \theta_{z}=\dfrac{1}{\sqrt{2}}\\ \\ \mathrm{sen\,}\theta_{z}=\dfrac{-1}{\sqrt{2}} \end{array}\right. \Rightarrow \boxed{\theta_{z}=\dfrac{7\pi}{ 4}}$


e) $z=4i$

$a=0\\ \\ b=4\\ \\ \\ \left|z\right|=\sqrt{\left(0\right)^{2}+\left(4\right)^{2}}\\ \\ \left|z\right|=\sqrt{0+16}\\ \\ \boxed{\left|z\right|=4}\\ \\ \\ \left\{ \begin{array}{l} \cos \theta_{z}=\dfrac{0}{4}=0\\ \\ \mathrm{sen\,}\theta_{z}=\dfrac{4}{4}=1 \end{array}\right. \Rightarrow \boxed{\theta_{z}=\dfrac{\pi}{2}}$


f) $z=2+2\sqrt{3}i$

$a=2\\ \\ b=2\sqrt{3}\\ \\ \\ \left|z\right|=\sqrt{\left(2\right)^{2}+\left(2\sqrt{3}\right)^{2}}\\ \\ \left|z\right|=\sqrt{4+12}\\ \\ \left|z\right|=\sqrt{16}\\ \\ \boxed{\left|z\right|=4}\\ \\ \\ \left\{ \begin{array}{l} \cos \theta_{z}=\dfrac{2}{4}=\dfrac{1}{2}\\ \\ \mathrm{sen\,}\theta_{z}=\dfrac{2\sqrt{3}}{4}=\dfrac{\sqrt{3}}{2} \end{array}\right. \Rightarrow \boxed{\theta_{z}=\dfrac{\pi}{6}}$