📌 Resumo teórico
Forma trigonométrica
\[ z=\rho\,\mathrm{e}^{\mathrm{i}\theta}=\rho\,(\cos\theta+\mathrm{i}\sin\theta),\qquad \rho=|z|\gt 0 \]
\(\theta\) é um argumento de \(z\); os outros são \(\theta+2k\pi\), \(k\in\mathbb{Z}\). O argumento principal pertence a \(]-\pi,\pi]\).
| De algébrica para trigonométrica | De trigonométrica para algébrica |
|---|---|
| \(\rho=\sqrt{a^{2}+b^{2}}\); \(\ \tan\theta=\dfrac{b}{a}\), com \(\theta\) no quadrante do afixo | \(a=\rho\cos\theta\), \(\ b=\rho\sin\theta\) |
Sobre os eixos: \(\rho=\rho\,\mathrm{e}^{\mathrm{i}0}\), \(-\rho=\rho\,\mathrm{e}^{\mathrm{i}\pi}\), \(\rho\mathrm{i}=\rho\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}\), \(-\rho\mathrm{i}=\rho\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\).
Operações na forma trigonométrica
\[ \rho_1\mathrm{e}^{\mathrm{i}\theta_1}\times\rho_2\mathrm{e}^{\mathrm{i}\theta_2}=\rho_1\rho_2\,\mathrm{e}^{\mathrm{i}(\theta_1+\theta_2)}\qquad \frac{\rho_1\mathrm{e}^{\mathrm{i}\theta_1}}{\rho_2\mathrm{e}^{\mathrm{i}\theta_2}}=\frac{\rho_1}{\rho_2}\,\mathrm{e}^{\mathrm{i}(\theta_1-\theta_2)}\qquad \left(\rho\,\mathrm{e}^{\mathrm{i}\theta}\right)^{n}=\rho^{n}\mathrm{e}^{\mathrm{i}n\theta} \]
Conjugado e simétrico
\(\overline{\rho\,\mathrm{e}^{\mathrm{i}\theta}}=\rho\,\mathrm{e}^{-\mathrm{i}\theta}\) (reflexão no eixo real); \(\ -\rho\,\mathrm{e}^{\mathrm{i}\theta}=\rho\,\mathrm{e}^{\mathrm{i}(\theta+\pi)}\) (reflexão na origem).
Igualdade
\(\rho_1\mathrm{e}^{\mathrm{i}\theta_1}=\rho_2\mathrm{e}^{\mathrm{i}\theta_2}\iff\rho_1=\rho_2\ \wedge\ \theta_1-\theta_2=2k\pi,\ k\in\mathbb{Z}\). Para que \(z\) seja real, um argumento tem de ser \(k\pi\); para que seja imaginário puro, \(\frac{\pi}{2}+k\pi\).
Na forma algébrica: \(\mathrm{i}^{n}=\mathrm{i}^{r}\), com \(r\) o resto da divisão de \(n\) por \(4\); dividir multiplicando pelo conjugado do denominador.
1. Demonstrações
1
Prove que:
1.1.
$\dfrac{2}{3}-\dfrac{1}{3}\,\mathrm{i}\;$ é solução da equação $\;1+2\mathrm{i}=3\mathrm{i}\,z\;$;
1.2.
$w=\dfrac{(1-3\mathrm{i})(2+\mathrm{i})}{1+2\,\mathrm{i}^{4n+1}}-\dfrac{3}{\mathrm{i}}\;$ é um número real;
1.3.
$\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}\,\mathrm{i}\;$ é solução da equação $\;z^{3}-z(3z-4)=0\;$;
1.4.
$z=\dfrac{(4-3\mathrm{i})^{2}}{\mathrm{i}}-\big(2\sqrt{6}\,\mathrm{i}\big)^{2}\;$ é imaginário puro.
1.1. Provar que $\dfrac{2}{3}-\dfrac{1}{3}\mathrm{i}$ é solução de $1+2\mathrm{i}=3\mathrm{i}\,z$.
Passo 1 Substituir no 2.º membro:
$$3\mathrm{i}\!\left(\tfrac{2}{3}-\tfrac{1}{3}\mathrm{i}\right)=3\mathrm{i}\cdot\tfrac{2}{3}-3\mathrm{i}\cdot\tfrac{1}{3}\mathrm{i}=2\mathrm{i}-\mathrm{i}^{2}=2\mathrm{i}+1=1+2\mathrm{i}\,.$$
O 2.º membro coincide com o 1.º $\Rightarrow$ a igualdade verifica-se. C.Q.D.
1.2. Provar que $w=\dfrac{(1-3\mathrm{i})(2+\mathrm{i})}{1+2\mathrm{i}^{4n+1}}-\dfrac{3}{\mathrm{i}}$ é real.
Passo 1 Como $\mathrm{i}^{4n}=1$, vem $\mathrm{i}^{4n+1}=\mathrm{i}$, logo $1+2\mathrm{i}^{4n+1}=1+2\mathrm{i}$.
Passo 2 Calcular o numerador:
$$(1-3\mathrm{i})(2+\mathrm{i})=2+\mathrm{i}-6\mathrm{i}-3\mathrm{i}^{2}=2-5\mathrm{i}+3=5-5\mathrm{i}.$$
Passo 3 Racionalizar:
$$\frac{5-5\mathrm{i}}{1+2\mathrm{i}}\cdot\frac{1-2\mathrm{i}}{1-2\mathrm{i}}=\frac{5-10\mathrm{i}-5\mathrm{i}+10\mathrm{i}^{2}}{1+4}=\frac{-5-15\mathrm{i}}{5}=-1-3\mathrm{i}.$$
Passo 4 Como $\dfrac{3}{\mathrm{i}}=\dfrac{3\,(-\mathrm{i})}{1}=-3\mathrm{i}$, então
$$w=(-1-3\mathrm{i})-(-3\mathrm{i})=-1.$$
$w=-1\in\mathbb{R}$. C.Q.D.
1.3. Provar que $z=\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}\mathrm{i}$ é solução de $z^{3}-z(3z-4)=0$.
Passo 1 Equação equivalente: $z^{3}-3z^{2}+4z=0\;\Leftrightarrow\;z(z^{2}-3z+4)=0$.
Passo 2 Calcular $z^{2}$:
$$z^{2}=\left(\tfrac{3}{2}-\tfrac{\sqrt{7}}{2}\mathrm{i}\right)^{2}=\tfrac{9}{4}-2\cdot\tfrac{3}{2}\cdot\tfrac{\sqrt{7}}{2}\mathrm{i}+\tfrac{7}{4}\mathrm{i}^{2}=\tfrac{9-7}{4}-\tfrac{3\sqrt{7}}{2}\mathrm{i}=\tfrac{1}{2}-\tfrac{3\sqrt{7}}{2}\mathrm{i}.$$
Passo 3 Calcular $z^{2}-3z+4$:
$$\tfrac{1}{2}-\tfrac{3\sqrt{7}}{2}\mathrm{i}-3\!\left(\tfrac{3}{2}-\tfrac{\sqrt{7}}{2}\mathrm{i}\right)+4=\tfrac{1}{2}-\tfrac{9}{2}+4-\tfrac{3\sqrt{7}}{2}\mathrm{i}+\tfrac{3\sqrt{7}}{2}\mathrm{i}=0.$$
Logo $z(z^{2}-3z+4)=z\cdot 0=0$. C.Q.D.
1.4. Provar que $z=\dfrac{(4-3\mathrm{i})^{2}}{\mathrm{i}}-(2\sqrt{6}\,\mathrm{i})^{2}$ é imaginário puro.
Passo 1 $(4-3\mathrm{i})^{2}=16-24\mathrm{i}+9\mathrm{i}^{2}=7-24\mathrm{i}$.
Passo 2 Dividir por $\mathrm{i}$ (multiplicar por $-\mathrm{i}$):
$$\frac{7-24\mathrm{i}}{\mathrm{i}}=(7-24\mathrm{i})(-\mathrm{i})=-7\mathrm{i}+24\mathrm{i}^{2}=-24-7\mathrm{i}.$$
Passo 3 $(2\sqrt{6}\,\mathrm{i})^{2}=4\cdot 6\cdot\mathrm{i}^{2}=-24$.
Passo 4 $z=(-24-7\mathrm{i})-(-24)=-7\mathrm{i}$.
$\operatorname{Re}(z)=0$ $\Rightarrow$ $z$ é imaginário puro. C.Q.D.
2. Forma trigonométrica
2
Escreva na forma trigonométrica os seguintes números complexos.
2.1.
$z=4\mathrm{i}$
2.2.
$z=-3$
2.3.
$z=1-\mathrm{i}$
2.4.
$z=\sqrt{3}-\mathrm{i}$
2.5.
$z=-2+2\sqrt{3}\,\mathrm{i}$
2.6.
$z=-4-4\mathrm{i}$
2.7.
$z=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\mathrm{i}$
2.8.
$z=\dfrac{(3+\mathrm{i})^{2}-4+2\mathrm{i}}{2-\mathrm{i}}$
2.9.
$z=2\mathrm{i}\,(1+\mathrm{i})^{-3}$
Recorda que: para $z=a+b\mathrm{i}$, $\;|z|=\sqrt{a^{2}+b^{2}}\;$ e o argumento é determinado por $\tan\theta=\dfrac{b}{a}$, escolhendo $\theta$ no quadrante da imagem de $z$.
2.1. $z=4\mathrm{i}$
Imaginário puro positivo: $|z|=4$ e $\arg(z)=\dfrac{\pi}{2}$.
$z=4\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$
2.2. $z=-3$
Real negativo: $|z|=3$ e $\arg(z)=\pi$.
$z=3\,\mathrm{e}^{\mathrm{i}\pi}$
2.3. $z=1-\mathrm{i}$
$|z|=\sqrt{1+1}=\sqrt{2}$. Imagem no 4.º quadrante: $\tan\theta=\dfrac{-1}{1}=-1\Rightarrow\theta=-\dfrac{\pi}{4}$.
$z=\sqrt{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}$
2.4. $z=\sqrt{3}-\mathrm{i}$
$|z|=\sqrt{3+1}=2$. 4.º quadrante: $\tan\theta=\dfrac{-1}{\sqrt{3}}\Rightarrow\theta=-\dfrac{\pi}{6}$.
$z=2\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}$
2.5. $z=-2+2\sqrt{3}\,\mathrm{i}$
$|z|=\sqrt{4+12}=4$. 2.º quadrante: $\tan\alpha=\sqrt{3}\Rightarrow\alpha=\dfrac{\pi}{3}$, logo $\theta=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}$.
$z=4\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$
2.6. $z=-4-4\mathrm{i}$
$|z|=\sqrt{16+16}=4\sqrt{2}$. 3.º quadrante: $\tan\alpha=1\Rightarrow\alpha=\dfrac{\pi}{4}$, logo $\theta=-\pi+\dfrac{\pi}{4}=-\dfrac{3\pi}{4}$.
$z=4\sqrt{2}\,\mathrm{e}^{-\mathrm{i}\frac{3\pi}{4}}$
2.7. $z=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\mathrm{i}$
$|z|=\sqrt{\tfrac{1}{4}+\tfrac{3}{4}}=1$. 4.º quadrante: $\cos\theta=\tfrac{1}{2},\;\sin\theta=-\tfrac{\sqrt{3}}{2}\Rightarrow\theta=-\dfrac{\pi}{3}$.
$z=\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}$
2.8. $z=\dfrac{(3+\mathrm{i})^{2}-4+2\mathrm{i}}{2-\mathrm{i}}$
Passo 1 $(3+\mathrm{i})^{2}=9+6\mathrm{i}+\mathrm{i}^{2}=8+6\mathrm{i}$, logo o numerador é $8+6\mathrm{i}-4+2\mathrm{i}=4+8\mathrm{i}$.
Passo 2 Racionalizar: $\dfrac{4+8\mathrm{i}}{2-\mathrm{i}}\cdot\dfrac{2+\mathrm{i}}{2+\mathrm{i}}=\dfrac{8+4\mathrm{i}+16\mathrm{i}-8}{5}=\dfrac{20\mathrm{i}}{5}=4\mathrm{i}$.
$z=4\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$
2.9. $z=2\mathrm{i}\,(1+\mathrm{i})^{-3}$
Passo 1 $(1+\mathrm{i})^{2}=2\mathrm{i}\Rightarrow(1+\mathrm{i})^{3}=2\mathrm{i}(1+\mathrm{i})=-2+2\mathrm{i}$.
Passo 2 $z=\dfrac{2\mathrm{i}}{-2+2\mathrm{i}}=\dfrac{\mathrm{i}}{-1+\mathrm{i}}\cdot\dfrac{-1-\mathrm{i}}{-1-\mathrm{i}}=\dfrac{-\mathrm{i}-\mathrm{i}^{2}}{2}=\dfrac{1-\mathrm{i}}{2}$.
Passo 3 $|z|=\dfrac{\sqrt{2}}{2}$, $\arg(z)=-\dfrac{\pi}{4}$.
$z=\dfrac{\sqrt{2}}{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}$
3. Módulo e argumento
3
Determine o módulo e um argumento dos seguintes números complexos.
3.1.
$(1+2\mathrm{i})^{3}+13$
3.2.
$2-2\sqrt{3}\,\mathrm{i}^{21}+\dfrac{1}{2}\!\left(8\mathrm{e}^{-\mathrm{i}\frac{5\pi}{3}}\right)$
3.3.
$\dfrac{(1-\sqrt{3}\,\mathrm{i})+(\mathrm{i}-\sqrt{3})}{\sqrt{2}-\sqrt{6}}$
3.4.
$\dfrac{4\,\mathrm{i}^{120}}{-1+\mathrm{i}^{3}\sqrt{3}}$
3.1. $(1+2\mathrm{i})^{3}+13$
Passo 1 $(1+2\mathrm{i})^{2}=1+4\mathrm{i}-4=-3+4\mathrm{i}$.
Passo 2 $(1+2\mathrm{i})^{3}=(1+2\mathrm{i})(-3+4\mathrm{i})=-3+4\mathrm{i}-6\mathrm{i}-8=-11-2\mathrm{i}$.
Passo 3 $-11-2\mathrm{i}+13=2-2\mathrm{i}$. Módulo: $\sqrt{4+4}=2\sqrt{2}$; 4.º quadrante: $\theta=-\dfrac{\pi}{4}$.
$|z|=2\sqrt{2}$ e $\arg(z)=-\dfrac{\pi}{4}$.
3.2. $2-2\sqrt{3}\,\mathrm{i}^{21}+\dfrac{1}{2}\!\left(8\mathrm{e}^{-\mathrm{i}\frac{5\pi}{3}}\right)$
Passo 1 $\mathrm{i}^{21}=\mathrm{i}^{4\cdot 5+1}=\mathrm{i}$, logo $-2\sqrt{3}\,\mathrm{i}^{21}=-2\sqrt{3}\,\mathrm{i}$.
Passo 2 $\dfrac{1}{2}\cdot 8\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{3}}=4\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}=4\!\left(\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}\mathrm{i}\right)=2+2\sqrt{3}\,\mathrm{i}$.
Passo 3 Soma: $2-2\sqrt{3}\,\mathrm{i}+2+2\sqrt{3}\,\mathrm{i}=4$.
$|z|=4$ e $\arg(z)=0$.
3.3. $\dfrac{(1-\sqrt{3}\,\mathrm{i})+(\mathrm{i}-\sqrt{3})}{\sqrt{2}-\sqrt{6}}$
Passo 1 Numerador: $(1-\sqrt{3})+(1-\sqrt{3})\mathrm{i}=(1-\sqrt{3})(1+\mathrm{i})$.
Passo 2 Denominador: $\sqrt{2}-\sqrt{6}=\sqrt{2}\,(1-\sqrt{3})$.
Passo 3 Quociente: $\dfrac{(1-\sqrt{3})(1+\mathrm{i})}{\sqrt{2}\,(1-\sqrt{3})}=\dfrac{1+\mathrm{i}}{\sqrt{2}}$. Módulo $=1$ e argumento $=\dfrac{\pi}{4}$.
$|z|=1$ e $\arg(z)=\dfrac{\pi}{4}$.
3.4. $\dfrac{4\mathrm{i}^{120}}{-1+\mathrm{i}^{3}\sqrt{3}}$
Passo 1 $\mathrm{i}^{120}=1$ e $\mathrm{i}^{3}=-\mathrm{i}$, logo o quociente é $\dfrac{4}{-1-\sqrt{3}\,\mathrm{i}}$.
Passo 2 $|-1-\sqrt{3}\,\mathrm{i}|=2$; 3.º quadrante $\Rightarrow$ argumento $-\dfrac{2\pi}{3}$, ou seja $-1-\sqrt{3}\,\mathrm{i}=2\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$.
Passo 3 $\dfrac{4}{2\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}}=2\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$.
$|z|=2$ e $\arg(z)=\dfrac{2\pi}{3}$.
4. Operações na forma trigonométrica
4
Em $\mathbb{C}$, conjunto dos números complexos, considere $z_{1}=2\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}$, $\;z_{2}=-3-\sqrt{27}\,\mathrm{i}\;$ e $\;z_{3}=\dfrac{4}{\mathrm{i}}$.
Escreva na forma trigonométrica:
4.1.
$z_{1}\times z_{2}$
4.2.
$2z_{1}\times \overline{z_{3}}$
4.3.
$-\dfrac{\overline{z_{2}}}{\frac{1}{2}z_{1}}$
4.4.
$\dfrac{-\overline{z_{3}}\times z_{1}}{z_{2}}$
4.5.
$-\dfrac{1}{3}\overline{z_{2}}\times \dfrac{1}{\overline{z_{3}}}$
4.6.
$z_{1}+z_{2}+2\sqrt{3}\,\mathrm{i}$
$z_{1}=2\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}$; $z_{2}=-3-\sqrt{27}\,\mathrm{i}=-3-3\sqrt{3}\,\mathrm{i}$; $z_{3}=\dfrac{4}{\mathrm{i}}=-4\mathrm{i}$.
Em forma trigonométrica: $|z_{2}|=\sqrt{9+27}=6$, 3.º quadrante $\Rightarrow z_{2}=6\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$; e $z_{3}=4\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}$.
Em forma trigonométrica: $|z_{2}|=\sqrt{9+27}=6$, 3.º quadrante $\Rightarrow z_{2}=6\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$; e $z_{3}=4\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}$.
4.1. $z_{1}\times z_{2}$
$$2\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}\cdot 6\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}=12\,\mathrm{e}^{-\mathrm{i}\pi}=12\,\mathrm{e}^{\mathrm{i}\pi}.$$
$z_{1}z_{2}=12\,\mathrm{e}^{\mathrm{i}\pi}$
4.2. $2z_{1}\times\overline{z_{3}}$
$\overline{z_{3}}=4\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$ e $2z_{1}=4\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}$.
$$4\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}\cdot 4\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}=16\,\mathrm{e}^{\mathrm{i}\!\left(\frac{\pi}{2}-\frac{\pi}{3}\right)}=16\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}.$$
$2z_{1}\overline{z_{3}}=16\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}$
4.3. $-\dfrac{\overline{z_{2}}}{\frac{1}{2}z_{1}}$
$\overline{z_{2}}=6\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$ e $\dfrac{1}{2}z_{1}=\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}$.
$$\frac{6\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}}{\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}}=6\,\mathrm{e}^{\mathrm{i}\!\left(\frac{2\pi}{3}+\frac{\pi}{3}\right)}=6\,\mathrm{e}^{\mathrm{i}\pi}.$$
Multiplicar por $-1=\mathrm{e}^{\mathrm{i}\pi}$: $\;6\mathrm{e}^{\mathrm{i}\pi}\cdot\mathrm{e}^{\mathrm{i}\pi}=6\mathrm{e}^{\mathrm{i}\,2\pi}=6\mathrm{e}^{\mathrm{i}\,0}$.
$=6\,\mathrm{e}^{\mathrm{i}\cdot 0}$
4.4. $\dfrac{-\overline{z_{3}}\times z_{1}}{z_{2}}$
$-\overline{z_{3}}=4\mathrm{e}^{\mathrm{i}\!\left(\frac{\pi}{2}+\pi\right)}=4\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}$.
$$\frac{4\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\cdot 2\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}}{6\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}}=\frac{8}{6}\,\mathrm{e}^{\mathrm{i}\!\left(-\frac{\pi}{2}-\frac{\pi}{3}+\frac{2\pi}{3}\right)}=\frac{4}{3}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}.$$
$=\dfrac{4}{3}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}$
4.5. $-\dfrac{1}{3}\overline{z_{2}}\times\dfrac{1}{\overline{z_{3}}}$
$\overline{z_{2}}=6\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$, $\overline{z_{3}}=4\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$.
$$-\tfrac{1}{3}\cdot 6\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}\cdot\tfrac{1}{4}\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}=-\tfrac{1}{2}\,\mathrm{e}^{\mathrm{i}\!\left(\frac{2\pi}{3}-\frac{\pi}{2}\right)}=-\tfrac{1}{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}.$$
Multiplicar por $-1=\mathrm{e}^{\mathrm{i}\pi}$: $\;\tfrac{1}{2}\mathrm{e}^{\mathrm{i}\!\left(\frac{\pi}{6}+\pi\right)}=\tfrac{1}{2}\mathrm{e}^{\mathrm{i}\frac{7\pi}{6}}=\tfrac{1}{2}\mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}}$.
$=\dfrac{1}{2}\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}}$
4.6. $z_{1}+z_{2}+2\sqrt{3}\,\mathrm{i}$
Passo 1 Passar $z_{1}$ a forma algébrica: $z_{1}=2\!\left(\cos\!\left(-\tfrac{\pi}{3}\right)+\mathrm{i}\sin\!\left(-\tfrac{\pi}{3}\right)\right)=1-\sqrt{3}\,\mathrm{i}$.
Passo 2 Somar: $(1-\sqrt{3}\,\mathrm{i})+(-3-3\sqrt{3}\,\mathrm{i})+2\sqrt{3}\,\mathrm{i}=-2-2\sqrt{3}\,\mathrm{i}$.
Passo 3 Módulo $\sqrt{4+12}=4$; 3.º quadrante, $\tan\alpha=\sqrt{3}\Rightarrow\alpha=\tfrac{\pi}{3}$, logo argumento $-\pi+\tfrac{\pi}{3}=-\tfrac{2\pi}{3}$.
$=4\,\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$
5. Conjugado e simétrico (forma algébrica)
5
Escreva na forma algébrica o conjugado e o simétrico dos seguintes números complexos.
5.1.
$z=2\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$
5.2.
$z=\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}-2\mathrm{e}^{-\mathrm{i}\pi}$
5.3.
$z=\dfrac{8\mathrm{i}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}}{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}$
5.4.
$z=\dfrac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\times(\sqrt{3}-\mathrm{i})}{\mathrm{i}^{7}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}}$
5.1. $z=2\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$
$z=2\!\left(\cos\!\left(-\tfrac{2\pi}{3}\right)+\mathrm{i}\sin\!\left(-\tfrac{2\pi}{3}\right)\right)=2\!\left(-\tfrac{1}{2}-\tfrac{\sqrt{3}}{2}\mathrm{i}\right)=-1-\sqrt{3}\,\mathrm{i}$.
$\overline{z}=-1+\sqrt{3}\,\mathrm{i}\;$;$\;-z=1+\sqrt{3}\,\mathrm{i}$.
5.2. $z=\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}-2\mathrm{e}^{-\mathrm{i}\pi}$
$\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}=\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\mathrm{i}\;$ e $\;\mathrm{e}^{-\mathrm{i}\pi}=-1$.
$z=\tfrac{\sqrt{2}}{2}-\tfrac{\sqrt{2}}{2}\mathrm{i}-(-2)=\!\left(2+\tfrac{\sqrt{2}}{2}\right)-\tfrac{\sqrt{2}}{2}\mathrm{i}$.
$\overline{z}=\!\left(2+\dfrac{\sqrt{2}}{2}\right)+\dfrac{\sqrt{2}}{2}\mathrm{i}\;$;$\;-z=\!\left(-2-\dfrac{\sqrt{2}}{2}\right)+\dfrac{\sqrt{2}}{2}\mathrm{i}$.
5.3. $z=\dfrac{8\mathrm{i}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}}{\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}$
Passo 1 $\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$, logo o numerador é $8\,\mathrm{e}^{\mathrm{i}\!\left(\frac{\pi}{2}+\frac{\pi}{6}\right)}=8\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$.
Passo 2 $z=8\,\mathrm{e}^{\mathrm{i}\!\left(\frac{2\pi}{3}+\frac{\pi}{4}\right)}=8\,\mathrm{e}^{\mathrm{i}\frac{11\pi}{12}}$.
Passo 3 Como $\tfrac{11\pi}{12}=\pi-\tfrac{\pi}{12}$:
$\cos\!\tfrac{11\pi}{12}=-\cos\!\tfrac{\pi}{12}=-\tfrac{\sqrt{6}+\sqrt{2}}{4},\;\;\sin\!\tfrac{11\pi}{12}=\sin\!\tfrac{\pi}{12}=\tfrac{\sqrt{6}-\sqrt{2}}{4}$.
Passo 4 $z=8\!\left(-\tfrac{\sqrt{6}+\sqrt{2}}{4}+\tfrac{\sqrt{6}-\sqrt{2}}{4}\mathrm{i}\right)=(-2\sqrt{6}-2\sqrt{2})+(2\sqrt{6}-2\sqrt{2})\mathrm{i}$.
$\overline{z}=(-2\sqrt{6}-2\sqrt{2})-(2\sqrt{6}-2\sqrt{2})\mathrm{i}\;$;
$-z=(2\sqrt{6}+2\sqrt{2})+(2\sqrt{2}-2\sqrt{6})\mathrm{i}$.
$-z=(2\sqrt{6}+2\sqrt{2})+(2\sqrt{2}-2\sqrt{6})\mathrm{i}$.
5.4. $z=\dfrac{\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\times(\sqrt{3}-\mathrm{i})}{\mathrm{i}^{7}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}}$
Passo 1 $\sqrt{3}-\mathrm{i}=2\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}$ e $\mathrm{i}^{7}=\mathrm{i}^{4}\cdot\mathrm{i}^{3}=-\mathrm{i}=\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}$.
Passo 2 Numerador: $\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\cdot 2\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}=2\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$. Denominador: $\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\cdot\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}$.
Passo 3 $z=2\,\mathrm{e}^{\mathrm{i}\!\left(-\frac{2\pi}{3}-\frac{\pi}{6}\right)}=2\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}}=2\!\left(-\tfrac{\sqrt{3}}{2}-\tfrac{1}{2}\mathrm{i}\right)=-\sqrt{3}-\mathrm{i}$.
$\overline{z}=-\sqrt{3}+\mathrm{i}\;$;$\;-z=\sqrt{3}+\mathrm{i}$.
6. $z_{1}=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$, $\;z_{2}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{12}}$, $\;z_{3}=2+2\mathrm{i}$
6
Em $\mathbb{C}$, conjunto dos números complexos, considere $z_{1}=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$, $\;z_{2}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{12}}\;$ e $\;z_{3}=2+2\mathrm{i}$.
6.1.
Calcule $z_{1}-\mathrm{i}$ na forma trigonométrica.
6.2.
Determine o valor de $\dfrac{z_{3}\times z_{2}-2}{\sqrt{3}\,\mathrm{i}^{5}}$. Apresente o resultado na forma algébrica.
6.1. Calcular $z_{1}-\mathrm{i}$ na forma trigonométrica.
Passo 1 $z_{1}=\sqrt{3}\!\left(\cos\tfrac{2\pi}{3}+\mathrm{i}\sin\tfrac{2\pi}{3}\right)=\sqrt{3}\!\left(-\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}\mathrm{i}\right)=-\tfrac{\sqrt{3}}{2}+\tfrac{3}{2}\mathrm{i}$.
Passo 2 $z_{1}-\mathrm{i}=-\tfrac{\sqrt{3}}{2}+\tfrac{1}{2}\mathrm{i}$. Módulo $\sqrt{\tfrac{3}{4}+\tfrac{1}{4}}=1$.
Passo 3 2.º quadrante; $\cos\theta=-\tfrac{\sqrt{3}}{2},\;\sin\theta=\tfrac{1}{2}\Rightarrow\theta=\tfrac{5\pi}{6}$.
$z_{1}-\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{5\pi}{6}}$
6.2. Calcular $\dfrac{z_{3}\,z_{2}-2}{\sqrt{3}\,\mathrm{i}^{5}}$ (forma algébrica).
Passo 1 $z_{3}=2+2\mathrm{i}=2\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$.
Passo 2 $z_{3}\,z_{2}=2\sqrt{2}\cdot\sqrt{2}\,\mathrm{e}^{\mathrm{i}\!\left(\frac{\pi}{4}+\frac{\pi}{12}\right)}=4\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}=4\!\left(\tfrac{1}{2}+\tfrac{\sqrt{3}}{2}\mathrm{i}\right)=2+2\sqrt{3}\,\mathrm{i}$.
Passo 3 Numerador: $2+2\sqrt{3}\,\mathrm{i}-2=2\sqrt{3}\,\mathrm{i}$. Denominador: $\mathrm{i}^{5}=\mathrm{i}\Rightarrow\sqrt{3}\,\mathrm{i}^{5}=\sqrt{3}\,\mathrm{i}$.
Passo 4 $\dfrac{2\sqrt{3}\,\mathrm{i}}{\sqrt{3}\,\mathrm{i}}=2$.
$=2$
7. $z_{1}=\sqrt{3}\,\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$, $\;z_{2}=\dfrac{\sqrt{2}}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}-2^{-1}$
7
Em $\mathbb{C}$, conjunto dos números complexos, considere $z_{1}=\sqrt{3}\,\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}\;$ e $\;z_{2}=\dfrac{\sqrt{2}}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}-2^{-1}$.
7.1.
Sem recorrer à calculadora, determine o valor de $\;\dfrac{\overline{z_{1}}}{\frac{\sqrt{3}}{2}+z_{1}}\;$. Apresente o resultado na forma trigonométrica.
7.2.
Determine o número complexo $w=\dfrac{-z_{1}\times z_{2}}{\mathrm{i}}$. Apresente o resultado na forma trigonométrica.
7.1. $\dfrac{\overline{z_{1}}}{\frac{\sqrt{3}}{2}+z_{1}}$ (forma trigonométrica).
Passo 1 Forma algébrica de $z_{1}$: $z_{1}=\sqrt{3}\!\left(-\tfrac{1}{2}-\tfrac{\sqrt{3}}{2}\mathrm{i}\right)=-\tfrac{\sqrt{3}}{2}-\tfrac{3}{2}\mathrm{i}$.
Passo 2 $\overline{z_{1}}=-\tfrac{\sqrt{3}}{2}+\tfrac{3}{2}\mathrm{i}=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$.
Passo 3 $\tfrac{\sqrt{3}}{2}+z_{1}=\tfrac{\sqrt{3}}{2}-\tfrac{\sqrt{3}}{2}-\tfrac{3}{2}\mathrm{i}=-\tfrac{3}{2}\mathrm{i}=\tfrac{3}{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}$.
Passo 4 Quociente: $\dfrac{\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}}{\tfrac{3}{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}}=\dfrac{2\sqrt{3}}{3}\,\mathrm{e}^{\mathrm{i}\!\left(\frac{2\pi}{3}+\frac{\pi}{2}\right)}=\dfrac{2\sqrt{3}}{3}\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{6}}=\dfrac{2\sqrt{3}}{3}\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}}$.
$=\dfrac{2\sqrt{3}}{3}\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}}$
7.2. $w=\dfrac{-z_{1}\,z_{2}}{\mathrm{i}}$ (forma trigonométrica).
Passo 1 Simplificar $z_{2}$: $\tfrac{\sqrt{2}}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}=\tfrac{\sqrt{2}}{2}\!\left(\tfrac{\sqrt{2}}{2}+\tfrac{\sqrt{2}}{2}\mathrm{i}\right)=\tfrac{1}{2}+\tfrac{1}{2}\mathrm{i}$, logo $z_{2}=\tfrac{1}{2}+\tfrac{1}{2}\mathrm{i}-\tfrac{1}{2}=\tfrac{1}{2}\mathrm{i}=\tfrac{1}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$.
Passo 2 $-z_{1}=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\!\left(-\frac{2\pi}{3}+\pi\right)}=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.
Passo 3 $-z_{1}\,z_{2}=\sqrt{3}\cdot\tfrac{1}{2}\,\mathrm{e}^{\mathrm{i}\!\left(\frac{\pi}{3}+\frac{\pi}{2}\right)}=\tfrac{\sqrt{3}}{2}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{6}}$.
Passo 4 $w=\dfrac{\tfrac{\sqrt{3}}{2}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{6}}}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}}=\tfrac{\sqrt{3}}{2}\,\mathrm{e}^{\mathrm{i}\!\left(\frac{5\pi}{6}-\frac{\pi}{2}\right)}=\tfrac{\sqrt{3}}{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.
$w=\dfrac{\sqrt{3}}{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$
8. $z=1+3\mathrm{i}$, $\;\theta=\arg(z)$
8
Em $\mathbb{C}$, conjunto dos números complexos, considere $z=1+3\mathrm{i}$. Seja $\theta$ um argumento de $z$.
8.1.
Exprima, na forma trigonométrica, em função de $\theta$, o produto de $\mathrm{i}$ pelo simétrico de $z$.
8.2.
Determine o valor exato de $\cos\!\left(\theta+\dfrac{\pi}{6}\right)$.
8.1. Produto de $\mathrm{i}$ pelo simétrico de $z$, em função de $\theta$.
Passo 1 $|z|=\sqrt{1+9}=\sqrt{10}$, logo $z=\sqrt{10}\,\mathrm{e}^{\mathrm{i}\theta}$.
Passo 2 Simétrico: $-z=\sqrt{10}\,\mathrm{e}^{\mathrm{i}(\theta+\pi)}$.
Passo 3 Como $\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$:
$$\mathrm{i}\cdot(-z)=\sqrt{10}\,\mathrm{e}^{\mathrm{i}\!\left(\theta+\pi+\frac{\pi}{2}\right)}=\sqrt{10}\,\mathrm{e}^{\mathrm{i}\!\left(\theta+\frac{3\pi}{2}\right)}.$$
$\mathrm{i}\cdot(-z)=\sqrt{10}\,\mathrm{e}^{\mathrm{i}\!\left(\frac{3\pi}{2}+\theta\right)}$
8.2. Valor exato de $\cos\!\left(\theta+\dfrac{\pi}{6}\right)$.
Passo 1 Como $z=1+3\mathrm{i}=\sqrt{10}(\cos\theta+\mathrm{i}\sin\theta)$:
$\cos\theta=\dfrac{1}{\sqrt{10}}=\dfrac{\sqrt{10}}{10}\;$ e $\;\sin\theta=\dfrac{3}{\sqrt{10}}=\dfrac{3\sqrt{10}}{10}$.
Passo 2 Fórmula da soma:
$$\cos\!\left(\theta+\tfrac{\pi}{6}\right)=\cos\theta\cos\tfrac{\pi}{6}-\sin\theta\sin\tfrac{\pi}{6}=\tfrac{\sqrt{10}}{10}\cdot\tfrac{\sqrt{3}}{2}-\tfrac{3\sqrt{10}}{10}\cdot\tfrac{1}{2}.$$
Passo 3 Reduzir: $\dfrac{\sqrt{30}}{20}-\dfrac{3\sqrt{10}}{20}=\dfrac{\sqrt{30}-3\sqrt{10}}{20}$.
$\cos\!\left(\theta+\dfrac{\pi}{6}\right)=\dfrac{\sqrt{30}-3\sqrt{10}}{20}$
9. $z_{1}=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\mathrm{i}$, $\;z_{2}=2+3\mathrm{i}$
9
Em $\mathbb{C}$, conjunto dos números complexos, considere $z_{1}=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\mathrm{i}\;$ e $\;z_{2}=2+3\mathrm{i}$.
9.1.
Determine, na forma algébrica, $\mathrm{e}^{-\mathrm{i}\pi}\times(z_{2})^{-3}$.
9.2.
Resolva, em $\mathbb{C}$, a equação $\;|z_{1}|-\overline{z_{2}}\times z=\mathrm{i}\,z$. Apresente a solução na forma trigonométrica.
9.1. $\mathrm{e}^{-\mathrm{i}\pi}\times(z_{2})^{-3}$ (forma algébrica).
Passo 1 $\mathrm{e}^{-\mathrm{i}\pi}=-1$.
Passo 2 Calcular $z_{2}^{3}$:
$z_{2}^{2}=(2+3\mathrm{i})^{2}=4+12\mathrm{i}-9=-5+12\mathrm{i}$;$z_{2}^{3}=z_{2}\cdot z_{2}^{2}=(2+3\mathrm{i})(-5+12\mathrm{i})=-10+24\mathrm{i}-15\mathrm{i}-36=-46+9\mathrm{i}$.
Passo 3 $(z_{2})^{-3}=\dfrac{1}{-46+9\mathrm{i}}\cdot\dfrac{-46-9\mathrm{i}}{-46-9\mathrm{i}}=\dfrac{-46-9\mathrm{i}}{2116+81}=\dfrac{-46-9\mathrm{i}}{2197}$.
Passo 4 Multiplicar por $-1$: $\dfrac{46+9\mathrm{i}}{2197}=\dfrac{46}{2197}+\dfrac{9}{2197}\mathrm{i}$.
$=\dfrac{46}{2197}+\dfrac{9}{2197}\mathrm{i}$
9.2. Resolver $|z_{1}|-\overline{z_{2}}\,z=\mathrm{i}\,z$ (forma trigonométrica).
Passo 1 $|z_{1}|=\sqrt{\tfrac{1}{4}+\tfrac{3}{4}}=1$ e $\overline{z_{2}}=2-3\mathrm{i}$.
Passo 2 Equação: $1-(2-3\mathrm{i})z=\mathrm{i}\,z\;\Leftrightarrow\;1=\mathrm{i}\,z+(2-3\mathrm{i})z=z(2-2\mathrm{i})$.
Passo 3 $z=\dfrac{1}{2-2\mathrm{i}}\cdot\dfrac{2+2\mathrm{i}}{2+2\mathrm{i}}=\dfrac{2+2\mathrm{i}}{8}=\dfrac{1+\mathrm{i}}{4}$.
Passo 4 $|z|=\dfrac{\sqrt{2}}{4}$ e $\arg(z)=\dfrac{\pi}{4}$.
$z=\dfrac{\sqrt{2}}{4}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$