1Raízes cúbicas de $\;64\,\mathrm{e}^{\mathrm{i}\frac{6\pi}{5}}$
Recorda: as raízes índice $n$ de $w=r\,\mathrm{e}^{\mathrm{i}\theta}$ são $\;z_{k}=\sqrt[n]{r}\,\mathrm{e}^{\mathrm{i}\frac{\theta+2k\pi}{n}}\;,\;k=0,1,\ldots,n-1$.
Passo 1 Identificar módulo e argumento de $w=64\,\mathrm{e}^{\mathrm{i}\frac{6\pi}{5}}$: $\;|w|=64\;,\;\arg(w)=\dfrac{6\pi}{5}$.
Passo 2 Módulo das raízes cúbicas: $\;\sqrt[3]{64}=4$.
Passo 3 Argumentos: $\;\dfrac{\frac{6\pi}{5}+2k\pi}{3}=\dfrac{2\pi}{5}+\dfrac{2k\pi}{3}\;,\;k\in\{0,1,2\}$.
$k=0$: $\dfrac{2\pi}{5}$; $\;\;k=1$: $\dfrac{2\pi}{5}+\dfrac{2\pi}{3}=\dfrac{16\pi}{15}$; $\;\;k=2$: $\dfrac{2\pi}{5}+\dfrac{4\pi}{3}=\dfrac{26\pi}{15}$.
$z_{0}=4\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{5}}\;;\;z_{1}=4\,\mathrm{e}^{\mathrm{i}\frac{16\pi}{15}}\;;\;z_{2}=4\,\mathrm{e}^{\mathrm{i}\frac{26\pi}{15}}$
2Raízes quartas de $\;-\dfrac{81}{2}+\dfrac{81\sqrt{3}}{2}\mathrm{i}$
Passo 1 Módulo: $\;|w|=\sqrt{\left(\tfrac{81}{2}\right)^{2}+\left(\tfrac{81\sqrt{3}}{2}\right)^{2}}=\sqrt{\tfrac{81^{2}}{4}(1+3)}=81$.
Passo 2 Argumento (imagem no $2.^{\mathrm{o}}$ quadrante): $\;\tan\alpha=\dfrac{\frac{81\sqrt{3}}{2}}{\frac{81}{2}}=\sqrt{3}\Rightarrow\alpha=\dfrac{\pi}{3}\;\Rightarrow\;\arg(w)=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}$.
Passo 3 Raízes quartas: $|z|=\sqrt[4]{81}=3$ e argumentos $\;\dfrac{\frac{2\pi}{3}+2k\pi}{4}=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\;$, $k\in\{0,1,2,3\}$.
Passo 4 Converter para a forma $a+b\mathrm{i}$, usando $z=3\,\mathrm{e}^{\mathrm{i}\theta}=3(\cos\theta+\mathrm{i}\sin\theta)$:
• $\theta=\dfrac{\pi}{6}$: $\;z_{0}=\dfrac{3\sqrt{3}}{2}+\dfrac{3}{2}\mathrm{i}$.
• $\theta=\dfrac{2\pi}{3}$: $\;z_{1}=-\dfrac{3}{2}+\dfrac{3\sqrt{3}}{2}\mathrm{i}$.
• $\theta=\dfrac{7\pi}{6}$: $\;z_{2}=-\dfrac{3\sqrt{3}}{2}-\dfrac{3}{2}\mathrm{i}$.
• $\theta=\dfrac{5\pi}{3}$: $\;z_{3}=\dfrac{3}{2}-\dfrac{3\sqrt{3}}{2}\mathrm{i}$.
$\dfrac{3\sqrt{3}}{2}+\dfrac{3}{2}\mathrm{i}\;,\;-\dfrac{3}{2}+\dfrac{3\sqrt{3}}{2}\mathrm{i}\;,\;-\dfrac{3\sqrt{3}}{2}-\dfrac{3}{2}\mathrm{i}\;,\;\dfrac{3}{2}-\dfrac{3\sqrt{3}}{2}\mathrm{i}$
3Equações com raízes em ℂ
3.1. $z^{3}=\dfrac{16}{z}$
Passo 1 Como $z\neq 0$, multiplica-se por $z$: $\;z^{4}=16=16\,\mathrm{e}^{\mathrm{i}\,0}$.
Passo 2 Raízes quartas: $|z|=2$, argumentos $\dfrac{2k\pi}{4}=\dfrac{k\pi}{2}$.
$z=2\,\mathrm{e}^{\mathrm{i}\,0}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\pi}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{2}}$
3.2. $z^{5}-z\,\mathrm{i}=0$
Passo 1 Pôr em evidência $z$: $\;z(z^{4}-\mathrm{i})=0$.
Passo 2 Logo $z=0\;\vee\;z^{4}=\mathrm{i}=\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}$.
Passo 3 As soluções não nulas são raízes quartas de $\mathrm{i}$: $|z|=1$ e $\arg=\dfrac{\frac{\pi}{2}+2k\pi}{4}=\dfrac{\pi}{8}+\dfrac{k\pi}{2}$.
$z=0\;,\;\mathrm{e}^{\mathrm{i}\frac{\pi}{8}}\;,\;\mathrm{e}^{\mathrm{i}\frac{5\pi}{8}}\;,\;\mathrm{e}^{\mathrm{i}\frac{9\pi}{8}}\;,\;\mathrm{e}^{\mathrm{i}\frac{13\pi}{8}}$
3.3. $z^{3}+64\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{5}}=0$
Passo 1 $z^{3}=-64\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{5}}=64\,\mathrm{e}^{\mathrm{i}\pi}\!\cdot\!\mathrm{e}^{\mathrm{i}\frac{3\pi}{5}}=64\,\mathrm{e}^{\mathrm{i}\frac{8\pi}{5}}$.
Passo 2 $|z|=\sqrt[3]{64}=4$ e $\arg=\dfrac{\frac{8\pi}{5}+2k\pi}{3}=\dfrac{8\pi}{15}+\dfrac{2k\pi}{3}$.
$k=0$: $\dfrac{8\pi}{15}$; $\;k=1$: $\dfrac{8\pi+10\pi}{15}=\dfrac{18\pi}{15}=\dfrac{6\pi}{5}$; $\;k=2$: $\dfrac{8\pi+20\pi}{15}=\dfrac{28\pi}{15}$.
$z=4\,\mathrm{e}^{\mathrm{i}\frac{8\pi}{15}}\;\vee\;z=4\,\mathrm{e}^{\mathrm{i}\frac{6\pi}{5}}\;\vee\;z=4\,\mathrm{e}^{\mathrm{i}\frac{28\pi}{15}}$
3.4. $(z^{2}+9)(z^{3}-8)=0$
Passo 1 Lei do anulamento: $\;z^{2}=-9\;\vee\;z^{3}=8$.
Passo 2 $z^{2}=-9=9\,\mathrm{e}^{\mathrm{i}\pi}\;\Rightarrow\;|z|=3$, $\arg=\dfrac{\pi}{2}+k\pi$ $\Rightarrow$ $z=\pm 3\mathrm{i}=3\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}},\;3\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{2}}$.
Passo 3 $z^{3}=8=8\,\mathrm{e}^{\mathrm{i}\,0}\;\Rightarrow\;|z|=2$, $\arg=\dfrac{2k\pi}{3}$ $\Rightarrow$ $z=2\,\mathrm{e}^{\mathrm{i}\,0},\;2\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}},\;2\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}$.
$z\in\left\{3\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}},\,3\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{2}},\,2\,\mathrm{e}^{\mathrm{i}\,0},\,2\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}},\,2\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}\right\}$
4Raízes quadradas de $-21+20\,\mathrm{i}$ na forma $a+b\mathrm{i}$
Passo 1 Procurar $a,b\in\mathbb{R}$ tais que $(a+b\mathrm{i})^{2}=-21+20\mathrm{i}$.
Passo 2 $(a+b\mathrm{i})^{2}=a^{2}-b^{2}+2ab\,\mathrm{i}$. Igualar partes real e imaginária:
$$\begin{cases}a^{2}-b^{2}=-21\\ 2ab=20\;\;\Leftrightarrow\;\;ab=10\end{cases}$$
Passo 3 Da 2.ª equação $b=\dfrac{10}{a}$. Substituindo: $\;a^{2}-\dfrac{100}{a^{2}}=-21\;\Leftrightarrow\;a^{4}+21a^{2}-100=0$.
Passo 4 Fazendo $t=a^{2}$: $\;t^{2}+21t-100=0\;\Rightarrow\;t=\dfrac{-21\pm\sqrt{441+400}}{2}=\dfrac{-21\pm 29}{2}$.
Como $t=a^{2}\geq 0$, vem $t=4$, ou seja $a=\pm 2$.
Passo 5 Para $a=2\Rightarrow b=5$; para $a=-2\Rightarrow b=-5$.
As raízes quadradas são $\;2+5\mathrm{i}\;$ e $\;-2-5\mathrm{i}$.
5Restantes raízes sextas de $w$
As $n$ raízes índice $n$ de $w$ estão em progressão geométrica de razão $\mathrm{e}^{\mathrm{i}\frac{2\pi}{n}}$ (ver Ex. 9).
Passo 1 Para $n=6$ a razão é $\mathrm{e}^{\mathrm{i}\frac{2\pi}{6}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.
Passo 2 Partindo de $z_{0}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{\pi}{5}}$, as restantes obtêm-se somando sucessivamente $\dfrac{\pi}{3}=\dfrac{5\pi}{15}$ ao argumento.
• $z_{1}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{5}+\frac{\pi}{3}\right)}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{8\pi}{15}}$
• $z_{2}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{5}+\frac{2\pi}{3}\right)}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{13\pi}{15}}$
• $z_{3}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{5}+\pi\right)}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{6\pi}{5}}$
• $z_{4}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{5}+\frac{4\pi}{3}\right)}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{23\pi}{15}}$
• $z_{5}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{5}+\frac{5\pi}{3}\right)}=\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{28\pi}{15}}$
$\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{8\pi}{15}}\,,\;\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{13\pi}{15}}\,,\;\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{6\pi}{5}}\,,\;\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{23\pi}{15}}\,,\;\dfrac{1}{3}\mathrm{e}^{\mathrm{i}\frac{28\pi}{15}}$
6Raízes quadradas de $\;\dfrac{\overline{w}}{-4w}$
Passo 1 $w=8\,\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}\;\Rightarrow\;\overline{w}=8\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$.
Passo 2 $-4w=4\,\mathrm{e}^{\mathrm{i}\pi}\!\cdot 8\,\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}=32\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.
Passo 3 $\dfrac{\overline{w}}{-4w}=\dfrac{8\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}}{32\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}}=\dfrac{1}{4}\,\mathrm{e}^{\mathrm{i}\left(\frac{2\pi}{3}-\frac{\pi}{3}\right)}=\dfrac{1}{4}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.
Passo 4 Raízes quadradas: $|z|=\dfrac{1}{2}$ e $\arg=\dfrac{\pi}{6}+k\pi$.
• $\theta=\dfrac{\pi}{6}$: $\;z=\dfrac{1}{2}\!\left(\cos\dfrac{\pi}{6}+\mathrm{i}\sin\dfrac{\pi}{6}\right)=\dfrac{\sqrt{3}}{4}+\dfrac{1}{4}\mathrm{i}$.
• $\theta=\dfrac{7\pi}{6}$: $\;z=-\dfrac{\sqrt{3}}{4}-\dfrac{1}{4}\mathrm{i}$.
$\dfrac{1}{2}\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}=\dfrac{\sqrt{3}}{4}+\dfrac{1}{4}\mathrm{i}\;\;;\;\;\dfrac{1}{2}\mathrm{e}^{\mathrm{i}\frac{7\pi}{6}}=-\dfrac{\sqrt{3}}{4}-\dfrac{1}{4}\mathrm{i}$
8$z_{1}=\mathrm{e}^{\mathrm{i}\frac{11\pi}{12}}$, $\;z_{2}=\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}$, $\;z_{3}=\sqrt{2}+\sqrt{2}\mathrm{i}$
8.1. $\dfrac{z_{1}}{z_{2}}$
Passo 1 Quociente em forma exponencial: $\;\dfrac{z_{1}}{z_{2}}=\mathrm{e}^{\mathrm{i}\left(\frac{11\pi}{12}-\frac{\pi}{6}\right)}=\mathrm{e}^{\mathrm{i}\frac{11\pi-2\pi}{12}}=\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}$.
Passo 2 Em forma algébrica: $\;\cos\dfrac{3\pi}{4}+\mathrm{i}\sin\dfrac{3\pi}{4}=-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\mathrm{i}$.
$\dfrac{z_{1}}{z_{2}}=\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}=-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\mathrm{i}$
8.2. $z_{2}$ na forma $a+b\mathrm{i}$
$z_{2}=\cos\dfrac{\pi}{6}+\mathrm{i}\sin\dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\mathrm{i}$.
$z_{2}=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\mathrm{i}$
8.3. $\dfrac{z_{3}}{z_{2}}$ na forma $a+b\mathrm{i}$
Passo 1 Forma trigonométrica de $z_{3}$: $\;|z_{3}|=\sqrt{2+2}=2$, $\arg=\dfrac{\pi}{4}\;\Rightarrow\;z_{3}=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$.
Passo 2 $\dfrac{z_{3}}{z_{2}}=2\,\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{4}-\frac{\pi}{6}\right)}=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{12}}$.
Passo 3 Usando $\cos\dfrac{\pi}{12}=\dfrac{\sqrt{6}+\sqrt{2}}{4}$ e $\sin\dfrac{\pi}{12}=\dfrac{\sqrt{6}-\sqrt{2}}{4}$:
$$\frac{z_{3}}{z_{2}}=2\!\left(\frac{\sqrt{6}+\sqrt{2}}{4}+\frac{\sqrt{6}-\sqrt{2}}{4}\mathrm{i}\right)=\frac{\sqrt{6}+\sqrt{2}}{2}+\frac{\sqrt{6}-\sqrt{2}}{2}\mathrm{i}.$$
$\dfrac{z_{3}}{z_{2}}=\dfrac{\sqrt{6}+\sqrt{2}}{2}+\dfrac{\sqrt{6}-\sqrt{2}}{2}\mathrm{i}$
8.4. $z=\dfrac{z_{1}}{z_{2}}+\dfrac{z_{3}}{z_{2}}$
Somando os resultados de 8.1 e 8.3:
$$\text{Re}(z)=-\frac{\sqrt{2}}{2}+\frac{\sqrt{6}+\sqrt{2}}{2}=\frac{\sqrt{6}}{2}\;,\;\;\text{Im}(z)=\frac{\sqrt{2}}{2}+\frac{\sqrt{6}-\sqrt{2}}{2}=\frac{\sqrt{6}}{2}.$$
$z=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{6}}{2}\mathrm{i}$
8.5. Determinar $w$ sabendo que $z$ é raiz quarta de $w$
Passo 1 $|z|=\sqrt{\dfrac{6}{4}+\dfrac{6}{4}}=\sqrt{3}\;$ e $\;\arg(z)=\dfrac{\pi}{4}$ (1.º quadrante, partes real e imaginária iguais e positivas), pelo que $z=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$.
Passo 2 $w=z^{4}=(\sqrt{3})^{4}\,\mathrm{e}^{\mathrm{i}\pi}=9\,\mathrm{e}^{\mathrm{i}\pi}=-9$.
$w=9\,\mathrm{e}^{\mathrm{i}\pi}=-9$
8.6. Área do triângulo $[OPQ]$
Afixos: $O=(0,0)$, $\;P=\left(\dfrac{\sqrt{6}}{2},\dfrac{\sqrt{6}}{2}\right)$, $\;Q=(-9,0)$.
Tomando $[OQ]$ como base, com $\overline{OQ}=9$, a altura correspondente é a distância de $P$ ao eixo real, isto é, $\dfrac{\sqrt{6}}{2}$.
$$\text{Área}=\frac{1}{2}\cdot 9\cdot\frac{\sqrt{6}}{2}=\frac{9\sqrt{6}}{4}.$$
Área $\;[OPQ]=\dfrac{9\sqrt{6}}{4}$
9Demonstrações sobre raízes índice $n$
$w=r\,\mathrm{e}^{\mathrm{i}\theta}$ não nulo, $n>1$. As raízes são $\;z_{k}=\sqrt[n]{r}\,\mathrm{e}^{\mathrm{i}\frac{\theta+2k\pi}{n}},\;k=0,\ldots,n-1$.
9.1. Progressão geométrica de razão $\mathrm{e}^{\mathrm{i}\frac{2\pi}{n}}$
Para $k\geq 1$: $\;\dfrac{z_{k}}{z_{k-1}}=\dfrac{\sqrt[n]{r}\,\mathrm{e}^{\mathrm{i}\frac{\theta+2k\pi}{n}}}{\sqrt[n]{r}\,\mathrm{e}^{\mathrm{i}\frac{\theta+2(k-1)\pi}{n}}}=\mathrm{e}^{\mathrm{i}\frac{2\pi}{n}}$.
A razão é constante, logo $(z_{k})$ é uma PG de razão $\mathrm{e}^{\mathrm{i}\frac{2\pi}{n}}$. C.Q.D.
9.2. $z_{0}+z_{1}+\ldots+z_{n-1}=0$
Soma de uma PG com razão $q=\mathrm{e}^{\mathrm{i}\frac{2\pi}{n}}\neq 1$:
$$S=z_{0}\cdot\frac{q^{n}-1}{q-1}\quad\text{e}\quad q^{n}=\mathrm{e}^{\mathrm{i}\,2\pi}=1.$$
Portanto $\;S=z_{0}\cdot\dfrac{0}{q-1}=0$.
$z_{0}+z_{1}+\ldots+z_{n-1}=0$. C.Q.D.
9.3. $z_{0}\,z_{1}\,\ldots\,z_{n-1}=(-1)^{n-1}\,w$
Passo 1 Como $z_{k}=z_{0}\,q^{k}$, o produto é $\;z_{0}^{\,n}\,q^{0+1+\ldots+(n-1)}=z_{0}^{\,n}\,q^{\frac{n(n-1)}{2}}$.
Passo 2 Por definição, $z_{0}^{\,n}=w$.
Passo 3 $q^{\frac{n(n-1)}{2}}=\mathrm{e}^{\mathrm{i}\frac{2\pi}{n}\cdot\frac{n(n-1)}{2}}=\mathrm{e}^{\mathrm{i}\,(n-1)\pi}=\left(\mathrm{e}^{\mathrm{i}\pi}\right)^{n-1}=(-1)^{n-1}$.
$z_{0}\,z_{1}\,\ldots\,z_{n-1}=(-1)^{n-1}\,w$. C.Q.D.
10$\sqrt{3}+\mathrm{i}\;$ é raiz índice 5 de $w$
10.1. Raízes quintas de $w$
Passo 1 Forma trigonométrica de $\sqrt{3}+\mathrm{i}$: $\;|\,.\,|=\sqrt{3+1}=2$ e $\arg=\arctan\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}$.
Logo $z_{0}=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}$ é uma das raízes quintas.
Passo 2 As restantes obtêm-se somando $\dfrac{2\pi}{5}=\dfrac{12\pi}{30}$ ao argumento; partindo de $\dfrac{\pi}{6}=\dfrac{5\pi}{30}$:
$\dfrac{5\pi}{30}\;,\;\dfrac{17\pi}{30}\;,\;\dfrac{29\pi}{30}\;,\;\dfrac{41\pi}{30}\;,\;\dfrac{53\pi}{30}$.
$2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}\,,\;2\,\mathrm{e}^{\mathrm{i}\frac{17\pi}{30}}\,,\;2\,\mathrm{e}^{\mathrm{i}\frac{29\pi}{30}}\,,\;2\,\mathrm{e}^{\mathrm{i}\frac{41\pi}{30}}\,,\;2\,\mathrm{e}^{\mathrm{i}\frac{53\pi}{30}}$
10.2. Perímetro do pentágono regular
Passo 1 Os afixos das raízes estão sobre uma circunferência de raio $R=2$.
Passo 2 O lado de um polígono regular de $n$ lados inscrito numa circunferência de raio $R$ é $\;\ell=2R\,\mathrm{sen}\,\dfrac{\pi}{n}$.
Para $n=5$ e $R=2$: $\;\ell=4\,\mathrm{sen}\dfrac{\pi}{5}$.
Passo 3 Perímetro $=5\ell=20\,\mathrm{sen}\dfrac{\pi}{5}$.
$P=20\,\mathrm{sen}\dfrac{\pi}{5}$
11Equações quadráticas em ℂ — fórmula resolvente
11.1. $z^{2}-6z+18=0$
$\Delta=36-72=-36\;\Rightarrow\;\sqrt{\Delta}=\pm 6\mathrm{i}$.
$z=\dfrac{6\pm 6\mathrm{i}}{2}=3\pm 3\mathrm{i}$.
$z=3+3\mathrm{i}\;\;\vee\;\;z=3-3\mathrm{i}$
11.2. $5z^{2}+2z+2=0$
$\Delta=4-40=-36\;\Rightarrow\;\sqrt{\Delta}=\pm 6\mathrm{i}$.
$z=\dfrac{-2\pm 6\mathrm{i}}{10}=-\dfrac{1}{5}\pm\dfrac{3}{5}\mathrm{i}$.
$z=-\dfrac{1}{5}+\dfrac{3}{5}\mathrm{i}\;\;\vee\;\;z=-\dfrac{1}{5}-\dfrac{3}{5}\mathrm{i}$
11.3. $z^{2}+\sqrt{3}\,z+7=0$
$\Delta=3-28=-25\;\Rightarrow\;\sqrt{\Delta}=\pm 5\mathrm{i}$.
$z=\dfrac{-\sqrt{3}\pm 5\mathrm{i}}{2}=-\dfrac{\sqrt{3}}{2}\pm\dfrac{5}{2}\mathrm{i}$.
$z=-\dfrac{\sqrt{3}}{2}+\dfrac{5}{2}\mathrm{i}\;\;\vee\;\;z=-\dfrac{\sqrt{3}}{2}-\dfrac{5}{2}\mathrm{i}$
14$\dfrac{z_{2}}{z_{1}^{4}}\;$ na forma trigonométrica
Passo 1 $z_{1}=1+\mathrm{i}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\;\Rightarrow\;z_{1}^{4}=(\sqrt{2})^{4}\,\mathrm{e}^{\mathrm{i}\pi}=4\,\mathrm{e}^{\mathrm{i}\pi}$.
Passo 2 $z_{2}=\dfrac{3}{2}+\dfrac{\sqrt{3}}{2}\mathrm{i}$: $\;|z_{2}|=\sqrt{\dfrac{9}{4}+\dfrac{3}{4}}=\sqrt{3}$, $\;\arg(z_{2})=\arctan\dfrac{\sqrt{3}/2}{3/2}=\arctan\dfrac{\sqrt{3}}{3}=\dfrac{\pi}{6}$.
Portanto $\;z_{2}=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{6}}$.
Passo 3 $\dfrac{z_{2}}{z_{1}^{4}}=\dfrac{\sqrt{3}}{4}\,\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{6}-\pi\right)}=\dfrac{\sqrt{3}}{4}\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{6}}=\dfrac{\sqrt{3}}{4}\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{6}}$ (somando $2\pi$).
$\dfrac{z_{2}}{z_{1}^{4}}=\dfrac{\sqrt{3}}{4}\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{6}}$
15$z=-1+\mathrm{i}$
15.1. Forma trigonométrica
$|z|=\sqrt{1+1}=\sqrt{2}$. Imagem no $2.^{\mathrm{o}}$ quadrante: $\tan\alpha=1\Rightarrow\alpha=\dfrac{\pi}{4}$, $\arg(z)=\pi-\dfrac{\pi}{4}=\dfrac{3\pi}{4}$.
$z=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}$
15.2. $(-1+\mathrm{i})^{4}$
$\;(-1+\mathrm{i})^{4}=\left(\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}\right)^{4}=4\,\mathrm{e}^{\mathrm{i}\,3\pi}=4\,\mathrm{e}^{\mathrm{i}\pi}=-4$.
$(-1+\mathrm{i})^{4}=-4$
15.3. Rotação de $\dfrac{\pi}{4}$ em torno de $O$
A rotação corresponde à multiplicação por $\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}$:
$$z'=z\cdot\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}\cdot\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\pi}=-\sqrt{2}.$$
$z'=-\sqrt{2}$
16$z=\dfrac{2}{1-\mathrm{i}}-2\mathrm{i}^{97}$
16.1. Formas algébrica e trigonométrica
Passo 1 Racionalizar: $\;\dfrac{2}{1-\mathrm{i}}=\dfrac{2(1+\mathrm{i})}{(1-\mathrm{i})(1+\mathrm{i})}=\dfrac{2(1+\mathrm{i})}{2}=1+\mathrm{i}$.
Passo 2 $\mathrm{i}^{97}=\mathrm{i}^{4\cdot 24+1}=\mathrm{i}\;\Rightarrow\;2\mathrm{i}^{97}=2\mathrm{i}$.
Passo 3 $z=(1+\mathrm{i})-2\mathrm{i}=1-\mathrm{i}$.
Passo 4 $|z|=\sqrt{2}$, imagem no $4.^{\mathrm{o}}$ quadrante com $\arg(z)=-\dfrac{\pi}{4}$.
$z=1-\mathrm{i}=\sqrt{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}$
16.2. Restantes raízes índice 3 de $w$
As três raízes índice 3 estão em PG de razão $\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$. Partindo de $z=\sqrt{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}$:
• $z_{1}=z\cdot\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{4}+\frac{2\pi}{3}\right)}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{12}}$.
• $z_{2}=z\cdot\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{4}+\frac{4\pi}{3}\right)}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{13\pi}{12}}$.
Geometricamente, os três afixos formam um triângulo equilátero inscrito numa circunferência de raio $\sqrt{2}$.
$\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{12}}\;$ e $\;\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{13\pi}{12}}$
17Raízes índice 8 de $1$
Passo 1 $1=\mathrm{e}^{\mathrm{i}\,0}$. As 8 raízes têm módulo $1$ e argumentos $\dfrac{2k\pi}{8}=\dfrac{k\pi}{4}\;,\;k=0,\ldots,7$.
$\mathrm{e}^{\mathrm{i}\,0}\,,\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\,,\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}\,,\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}\,,\,\mathrm{e}^{\mathrm{i}\pi}\,,\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}\,,\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{2}}\,,\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{4}}$.
Passo 2 — soma Aplicando o resultado do Ex. 9.2: a soma das raízes índice $n$ ($n>1$) é zero. Logo a soma é $0$.
Passo 3 — produto Aplicando o Ex. 9.3 com $w=1$ e $n=8$:
$$\prod_{k=0}^{7}z_{k}=(-1)^{8-1}\cdot 1=(-1)^{7}=-1.$$
Observação: a soma é, de facto, $0$. Quanto ao produto, do resultado geral resulta $(-1)^{n-1}=-1$ (pois $n=8$ é par). Pode confirmar-se diretamente: $\sum\arg=\dfrac{\pi}{4}(0+1+\ldots+7)=7\pi\;\Rightarrow\;\mathrm{e}^{\mathrm{i}\,7\pi}=-1$.
As 8 raízes são $\;\mathrm{e}^{\mathrm{i}\frac{k\pi}{4}},\;k=0,\ldots,7$. Soma $=0$ e produto $=-1$.
19$z_{1}=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$, $\;z_{2}=-\sqrt{2}-\sqrt{2}\mathrm{i}$, $\;z_{3}=2\,\mathrm{e}^{\mathrm{i}\theta}$
19.1. Argumento principal de $w$
Passo 1 $\overline{z_{1}}=2\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}\;\Rightarrow\;(\overline{z_{1}})^{5}=32\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{3}}$.
Passo 2 $|z_{1}|=2$ e $\;-z_{1}=2\,\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{3}+\pi\right)}=2\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}$.
Passo 3 Numerador: $\;32\,\mathrm{e}^{-\mathrm{i}\frac{5\pi}{3}}\cdot 2\cdot 2\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}=128\,\mathrm{e}^{\mathrm{i}\left(-\frac{5\pi}{3}+\frac{4\pi}{3}\right)}=128\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}$.
Passo 4 Denominador: $\;\dfrac{1}{2}z_{1}=\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}$.
Passo 5 $w=\dfrac{128\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{3}}}{\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}}=128\,\mathrm{e}^{-\mathrm{i}\frac{2\pi}{3}}$.
$\arg_{\mathrm{p}}(w)=-\dfrac{2\pi}{3}\;\;(\in\,]-\pi,\pi])$
19.2. Resolver $(z_{2})^{4}\times z_{1}=z\,\mathrm{i}$
Passo 1 $z_{2}=-\sqrt{2}-\sqrt{2}\mathrm{i}$: $|z_{2}|=2$, imagem no $3.^{\mathrm{o}}$ quadrante com $\arg=\dfrac{5\pi}{4}$. Logo $z_{2}=2\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}$.
Passo 2 $(z_{2})^{4}=2^{4}\,\mathrm{e}^{\mathrm{i}\,5\pi}=16\,\mathrm{e}^{\mathrm{i}\pi}=-16$.
Passo 3 $(z_{2})^{4}\cdot z_{1}=-16\cdot 2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}=32\,\mathrm{e}^{\mathrm{i}\pi}\!\cdot\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}=32\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}$.
Passo 4 $z=\dfrac{32\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}}{\mathrm{i}}=\dfrac{32\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}}{\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}}=32\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{6}}$.
Passo 5 Forma algébrica: $\;32\!\left(\cos\dfrac{5\pi}{6}+\mathrm{i}\sin\dfrac{5\pi}{6}\right)=32\!\left(-\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\mathrm{i}\right)=-16\sqrt{3}+16\mathrm{i}$.
$z=-16\sqrt{3}+16\mathrm{i}$
19.3. Valores de $\theta$ para $w=\dfrac{-2}{z_{2}\cdot(z_{3})^{3}}$ real
Passo 1 $z_{2}\cdot(z_{3})^{3}=2\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}\cdot 8\,\mathrm{e}^{\mathrm{i}\,3\theta}=16\,\mathrm{e}^{\mathrm{i}\left(\frac{5\pi}{4}+3\theta\right)}$.
Passo 2 $-2=2\,\mathrm{e}^{\mathrm{i}\pi}$. Logo $\;w=\dfrac{2\,\mathrm{e}^{\mathrm{i}\pi}}{16\,\mathrm{e}^{\mathrm{i}\left(\frac{5\pi}{4}+3\theta\right)}}=\dfrac{1}{8}\,\mathrm{e}^{\mathrm{i}\left(\pi-\frac{5\pi}{4}-3\theta\right)}=\dfrac{1}{8}\,\mathrm{e}^{\mathrm{i}\left(-\frac{\pi}{4}-3\theta\right)}$.
Passo 3 $w\in\mathbb{R}\;\Leftrightarrow\;\arg(w)=k\pi\,,\,k\in\mathbb{Z}$.
$$-\frac{\pi}{4}-3\theta=k\pi\;\Leftrightarrow\;3\theta=-\frac{\pi}{4}-k\pi\;\Leftrightarrow\;\theta=-\frac{\pi}{12}-\frac{k\pi}{3}.$$
$\theta=-\dfrac{\pi}{12}-\dfrac{k\pi}{3}\;,\;k\in\mathbb{Z}$
20$z_{1}=1+\mathrm{i}$, $\;z_{2}=\sqrt[4]{2}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$, $\;z_{3}=\sqrt[4]{2}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{3}}$
$z_{1}=1+\mathrm{i}=\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\;\Rightarrow\;\overline{z_{1}}=\sqrt{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}$.
20.1. Mostrar que $w=\dfrac{(z_{1})^{10}}{(\overline{z_{1}})^{6}}\in\mathbb{R}$
$(z_{1})^{10}=(\sqrt{2})^{10}\,\mathrm{e}^{\mathrm{i}\frac{10\pi}{4}}=32\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{2}}=32\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}=32\mathrm{i}$.
$(\overline{z_{1}})^{6}=(\sqrt{2})^{6}\,\mathrm{e}^{-\mathrm{i}\frac{6\pi}{4}}=8\,\mathrm{e}^{-\mathrm{i}\frac{3\pi}{2}}=8\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}=8\mathrm{i}$.
$w=\dfrac{32\mathrm{i}}{8\mathrm{i}}=4\in\mathbb{R}$.
$w=4$ é real. C.Q.D.
20.2. $\left(\dfrac{z_{1}}{\overline{z_{1}}}\right)^{n}$ imaginário puro de coeficiente negativo
Passo 1 $\dfrac{z_{1}}{\overline{z_{1}}}=\dfrac{\sqrt{2}\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}}{\sqrt{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{4}}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}\;\Rightarrow\;\left(\dfrac{z_{1}}{\overline{z_{1}}}\right)^{n}=\mathrm{e}^{\mathrm{i}\frac{n\pi}{2}}$.
Passo 2 Imaginário puro de coeficiente negativo $\Leftrightarrow$ argumento congruente com $-\dfrac{\pi}{2}$ (i.e. $\dfrac{3\pi}{2}$) módulo $2\pi$:
$$\dfrac{n\pi}{2}=\dfrac{3\pi}{2}+2k\pi\;\Leftrightarrow\;n=3+4k\,,\;k\in\mathbb{Z}.$$
Como $n\in\mathbb{N}$, $k\in\{0,1,2,\ldots\}$ (e $k$ inteiro tal que $n\geq 1$).
$n=3+4k\;,\;k\in\mathbb{Z}\;\;(n\in\mathbb{N})$
20.3. $a,b$ tais que $z_{1}^{5}+a\,z_{1}^{3}=-b$
$z_{1}^{3}=(\sqrt{2})^{3}\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}=2\sqrt{2}\!\left(-\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\mathrm{i}\right)=-2+2\mathrm{i}$.
$z_{1}^{5}=(\sqrt{2})^{5}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}=4\sqrt{2}\!\left(-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}\mathrm{i}\right)=-4-4\mathrm{i}$.
$z_{1}^{5}+a\,z_{1}^{3}=(-4-4\mathrm{i})+a(-2+2\mathrm{i})=(-4-2a)+(-4+2a)\mathrm{i}$.
Como $-b\in\mathbb{R}$, igualar a parte imaginária a $0$: $\;-4+2a=0\Rightarrow a=2$.
Parte real: $-4-2(2)=-8=-b\Rightarrow b=8$.
$a=2\;$ e $\;b=8$
20.4. $z_{2}$ e $z_{3}$ são raízes índice 4 do mesmo número
$(z_{2})^{4}=(\sqrt[4]{2})^{4}\,\mathrm{e}^{\mathrm{i}\frac{8\pi}{3}}=2\,\mathrm{e}^{\mathrm{i}\frac{8\pi}{3}}=2\,\mathrm{e}^{\mathrm{i}\left(\frac{8\pi}{3}-2\pi\right)}=2\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$.
$(z_{3})^{4}=2\,\mathrm{e}^{\mathrm{i}\frac{20\pi}{3}}=2\,\mathrm{e}^{\mathrm{i}\left(\frac{20\pi}{3}-6\pi\right)}=2\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$.
Como $(z_{2})^{4}=(z_{3})^{4}=2\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$, ambos são raízes índice 4 desse mesmo número.
O número complexo é $\;2\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}$. C.Q.D.
21$4z^{6}+256=0\;$ na forma algébrica
Passo 1 $z^{6}=-64=64\,\mathrm{e}^{\mathrm{i}\pi}$.
Passo 2 $|z|=\sqrt[6]{64}=2$ e $\arg=\dfrac{\pi+2k\pi}{6}=\dfrac{\pi}{6}+\dfrac{k\pi}{3}\,,\;k=0,\ldots,5$.
Passo 3 Converter cada raiz para a forma $a+b\mathrm{i}$:
• $\theta=\dfrac{\pi}{6}$: $\;2\!\left(\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\mathrm{i}\right)=\sqrt{3}+\mathrm{i}$.
• $\theta=\dfrac{\pi}{2}$: $\;2\mathrm{i}$.
• $\theta=\dfrac{5\pi}{6}$: $\;-\sqrt{3}+\mathrm{i}$.
• $\theta=\dfrac{7\pi}{6}$: $\;-\sqrt{3}-\mathrm{i}$.
• $\theta=\dfrac{3\pi}{2}$: $\;-2\mathrm{i}$.
• $\theta=\dfrac{11\pi}{6}$: $\;\sqrt{3}-\mathrm{i}$.
$z_{0}=\sqrt{3}+\mathrm{i}\;;\;z_{1}=2\mathrm{i}\;;\;z_{2}=-\sqrt{3}+\mathrm{i}\;;\;z_{3}=-\sqrt{3}-\mathrm{i}\;;\;z_{4}=-2\mathrm{i}\;;\;z_{5}=\sqrt{3}-\mathrm{i}$
22Equações $z^{n}=w$
22.1. $z^{2}=-16$
$-16=16\,\mathrm{e}^{\mathrm{i}\pi}\;\Rightarrow\;|z|=4,\;\arg=\dfrac{\pi}{2}+k\pi$.
$z=4\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}\;\vee\;z=4\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{2}}$
22.2. $z^{2}=4$
$4=4\,\mathrm{e}^{\mathrm{i}\,0}\;\Rightarrow\;|z|=2,\;\arg=k\pi$.
$z=2\,\mathrm{e}^{\mathrm{i}\times 0}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\pi}$
22.3. $z^{4}=-1$
$-1=\mathrm{e}^{\mathrm{i}\pi}\;\Rightarrow\;|z|=1,\;\arg=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$.
$z=\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\;\vee\;z=\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}\;\vee\;z=\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}\;\vee\;z=\mathrm{e}^{\mathrm{i}\frac{7\pi}{4}}$
22.4. $z^{5}=32\,\mathrm{e}^{\mathrm{i}\frac{\pi}{5}}$
$|z|=\sqrt[5]{32}=2$ e $\arg=\dfrac{\frac{\pi}{5}+2k\pi}{5}=\dfrac{\pi}{25}+\dfrac{2k\pi}{5}=\dfrac{\pi}{25}+\dfrac{10k\pi}{25}$.
$z=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{25}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{11\pi}{25}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{21\pi}{25}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{31\pi}{25}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{41\pi}{25}}$
22.5. $z^{3}=-3\sqrt{3}+9\mathrm{i}$
Passo 1 $|w|=\sqrt{27+81}=\sqrt{108}=6\sqrt{3}$. Imagem no $2.^{\mathrm{o}}$ quadrante: $\tan\alpha=\dfrac{9}{3\sqrt{3}}=\sqrt{3}\Rightarrow\alpha=\dfrac{\pi}{3}$, $\arg=\dfrac{2\pi}{3}$.
Passo 2 $|z|=\sqrt[3]{6\sqrt{3}}=\sqrt[6]{(6\sqrt{3})^{2}}=\sqrt[6]{108}$ e $\arg=\dfrac{2\pi}{9}+\dfrac{2k\pi}{3}$.
$z=\sqrt[6]{108}\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{9}}\;\vee\;z=\sqrt[6]{108}\,\mathrm{e}^{\mathrm{i}\frac{8\pi}{9}}\;\vee\;z=\sqrt[6]{108}\,\mathrm{e}^{\mathrm{i}\frac{14\pi}{9}}$
22.6. $z^{6}=\dfrac{128}{-1+\mathrm{i}}$
Passo 1 Racionalizar: $\;\dfrac{128}{-1+\mathrm{i}}\cdot\dfrac{-1-\mathrm{i}}{-1-\mathrm{i}}=\dfrac{128(-1-\mathrm{i})}{2}=-64-64\mathrm{i}$.
Passo 2 $|-64-64\mathrm{i}|=64\sqrt{2}$; imagem no $3.^{\mathrm{o}}$ quadrante com $\arg=-\dfrac{3\pi}{4}$.
Passo 3 $|z|=\sqrt[6]{64\sqrt{2}}=\sqrt[6]{2^{6}\cdot 2^{1/2}}=2\,\sqrt[12]{2}$ e $\arg=-\dfrac{\pi}{8}+\dfrac{k\pi}{3}$.
Para $k=0,1,2,3,4,5$ obtêm-se os argumentos $-\dfrac{\pi}{8}\,,\dfrac{5\pi}{24}\,,\dfrac{13\pi}{24}\,,\dfrac{7\pi}{8}\,,\dfrac{29\pi}{24}\,,\dfrac{37\pi}{24}$.
$z=2\sqrt[12]{2}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{8}}\;\vee\;z=2\sqrt[12]{2}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{24}}\;\vee\;z=2\sqrt[12]{2}\,\mathrm{e}^{\mathrm{i}\frac{13\pi}{24}}\;\vee$
$\vee\;z=2\sqrt[12]{2}\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{8}}\;\vee\;z=2\sqrt[12]{2}\,\mathrm{e}^{\mathrm{i}\frac{29\pi}{24}}\;\vee\;z=2\sqrt[12]{2}\,\mathrm{e}^{\mathrm{i}\frac{37\pi}{24}}$
23Cálculo e representação de raízes
Os afixos das raízes índice $n$ formam um polígono regular de $n$ vértices, inscrito numa circunferência de raio $\sqrt[n]{|w|}$ centrada na origem.
23.1. Raízes cúbicas de $-8\mathrm{i}$
$-8\mathrm{i}=8\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\;\Rightarrow\;|z|=2,\;\arg=-\dfrac{\pi}{6}+\dfrac{2k\pi}{3}$.
$z_{0}=2\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{6}}\;;\;z_{1}=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}\;;\;z_{2}=2\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{6}}$
23.2. Raízes quartas de $-256$
$-256=256\,\mathrm{e}^{\mathrm{i}\pi}\;\Rightarrow\;|z|=4,\;\arg=\dfrac{\pi}{4}+\dfrac{k\pi}{2}$.
$z_{0}=4\,\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\;;\;z_{1}=4\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{4}}\;;\;z_{2}=4\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}\;;\;z_{3}=4\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{4}}$
23.3. Raízes quintas de $-16\sqrt{2}+16\sqrt{2}\mathrm{i}$
$|w|=\sqrt{512+512}=32$. $2.^{\mathrm{o}}$ quadrante, $\tan\alpha=1\Rightarrow\arg=\dfrac{3\pi}{4}$.
$|z|=\sqrt[5]{32}=2$, $\arg=\dfrac{3\pi}{20}+\dfrac{2k\pi}{5}=\dfrac{3\pi}{20}+\dfrac{8k\pi}{20}$.
$z_{0}=2\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{20}}\;;\;z_{1}=2\,\mathrm{e}^{\mathrm{i}\frac{11\pi}{20}}\;;\;z_{2}=2\,\mathrm{e}^{\mathrm{i}\frac{19\pi}{20}}\;;\;z_{3}=2\,\mathrm{e}^{\mathrm{i}\frac{27\pi}{20}}\;;\;z_{4}=2\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{4}}$
23.4. Raízes sextas de $\dfrac{1}{\mathrm{i}}$
$\dfrac{1}{\mathrm{i}}=-\mathrm{i}=\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\;\Rightarrow\;|z|=1,\;\arg=-\dfrac{\pi}{12}+\dfrac{k\pi}{3}$.
$z_{0}=\mathrm{e}^{-\mathrm{i}\frac{\pi}{12}}\;;\;z_{1}=\mathrm{e}^{\mathrm{i}\frac{\pi}{4}}\;;\;z_{2}=\mathrm{e}^{\mathrm{i}\frac{7\pi}{12}}\;;\;z_{3}=\mathrm{e}^{\mathrm{i}\frac{11\pi}{12}}\;;\;z_{4}=\mathrm{e}^{\mathrm{i}\frac{5\pi}{4}}\;;\;z_{5}=\mathrm{e}^{\mathrm{i}\frac{19\pi}{12}}$
24$z_{1}=2\,\mathrm{e}^{\mathrm{i}\frac{\pi}{12}}\;$ é raiz quarta de $w$
24.1. As outras três raízes
A razão da PG é $\;\mathrm{e}^{\mathrm{i}\frac{2\pi}{4}}=\mathrm{e}^{\mathrm{i}\frac{\pi}{2}}\;$. Somando $\dfrac{\pi}{2}$ ao argumento sucessivamente:
• $z_{2}=2\,\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{12}+\frac{\pi}{2}\right)}=2\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{12}}$;
• $z_{3}=2\,\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{12}+\pi\right)}=2\,\mathrm{e}^{\mathrm{i}\frac{13\pi}{12}}$;
• $z_{4}=2\,\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{12}+\frac{3\pi}{2}\right)}=2\,\mathrm{e}^{\mathrm{i}\frac{19\pi}{12}}$.
$z_{2}=2\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{12}}\;;\;z_{3}=2\,\mathrm{e}^{\mathrm{i}\frac{13\pi}{12}}\;;\;z_{4}=2\,\mathrm{e}^{\mathrm{i}\frac{19\pi}{12}}$
24.2. Área do polígono
Os 4 afixos formam um quadrado inscrito numa circunferência de raio $R=2$.
As diagonais coincidem com diâmetros: $d=2R=4$. A área é $\;\dfrac{d^{2}}{2}=\dfrac{16}{2}=8$.
A medida da área é $\;8\;\mathrm{u.a.}$
24.3. Soma das quatro raízes
Pelo Ex. 9.2, a soma das raízes índice $n$ ($n>1$) de qualquer número não nulo é zero. Em particular, para $n=4$:
$$z_{1}+z_{2}+z_{3}+z_{4}=0.$$
A soma é $0$. C.Q.D.
25Equações em ℂ
25.1. $z^{4}-\mathrm{i}z+\sqrt{3}\,z=0$
Passo 1 Pôr $z$ em evidência: $\;z\!\left(z^{3}+\sqrt{3}-\mathrm{i}\right)=0$.
Passo 2 $z=0\;\vee\;z^{3}=\mathrm{i}-\sqrt{3}=-\sqrt{3}+\mathrm{i}$.
Passo 3 $|-\sqrt{3}+\mathrm{i}|=2$; imagem no $2.^{\mathrm{o}}$ quadrante, $\tan\alpha=\dfrac{1}{\sqrt{3}}\Rightarrow\arg=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}$.
Passo 4 $|z|=\sqrt[3]{2}$, $\arg=\dfrac{5\pi}{18}+\dfrac{2k\pi}{3}=\dfrac{5\pi}{18}+\dfrac{12k\pi}{18}$.
$z=0\;\vee\;z=\sqrt[3]{2}\,\mathrm{e}^{\mathrm{i}\frac{5\pi}{18}}\;\vee\;z=\sqrt[3]{2}\,\mathrm{e}^{\mathrm{i}\frac{17\pi}{18}}\;\vee\;z=\sqrt[3]{2}\,\mathrm{e}^{\mathrm{i}\frac{29\pi}{18}}$
25.2. $z^{5}+32\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}=0$
$z^{5}=-32\,\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}=32\,\mathrm{e}^{\mathrm{i}\pi}\!\cdot\mathrm{e}^{\mathrm{i}\frac{\pi}{3}}=32\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{3}}$.
$|z|=2$, $\arg=\dfrac{4\pi}{15}+\dfrac{2k\pi}{5}=\dfrac{4\pi}{15}+\dfrac{6k\pi}{15}$.
$k=0$: $\dfrac{4\pi}{15}$; $\;k=1$: $\dfrac{10\pi}{15}=\dfrac{2\pi}{3}$; $\;k=2$: $\dfrac{16\pi}{15}$; $\;k=3$: $\dfrac{22\pi}{15}$; $\;k=4$: $\dfrac{28\pi}{15}$.
$z=2\,\mathrm{e}^{\mathrm{i}\frac{4\pi}{15}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{2\pi}{3}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{16\pi}{15}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{22\pi}{15}}\;\vee\;z=2\,\mathrm{e}^{\mathrm{i}\frac{28\pi}{15}}$
25.3. $z^{3}+3\mathrm{i}\,\overline{z}=0$
Passo 1 Seja $z=r\,\mathrm{e}^{\mathrm{i}\theta}\,(r\geq 0)$. Então $\overline{z}=r\,\mathrm{e}^{-\mathrm{i}\theta}$.
Passo 2 A equação dá $\;r^{3}\,\mathrm{e}^{\mathrm{i}\,3\theta}=-3\mathrm{i}\cdot r\,\mathrm{e}^{-\mathrm{i}\theta}=3r\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\!\cdot\mathrm{e}^{-\mathrm{i}\theta}=3r\,\mathrm{e}^{-\mathrm{i}\left(\theta+\frac{\pi}{2}\right)}$.
Passo 3 Igualar módulos: $r^{3}=3r$. Se $r\neq 0$, $r^{2}=3\Rightarrow r=\sqrt{3}$. Caso contrário, $z=0$.
Passo 4 Igualar argumentos: $\;3\theta=-\theta-\dfrac{\pi}{2}+2k\pi\;\Leftrightarrow\;4\theta=-\dfrac{\pi}{2}+2k\pi\;\Leftrightarrow\;\theta=-\dfrac{\pi}{8}+\dfrac{k\pi}{2}$.
Os 4 valores distintos: $-\dfrac{\pi}{8},\;\dfrac{3\pi}{8},\;\dfrac{7\pi}{8},\;\dfrac{11\pi}{8}$.
$z=0\;\vee\;z=\sqrt{3}\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{8}}\;\vee\;z=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{3\pi}{8}}\;\vee\;z=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{7\pi}{8}}\;\vee\;z=\sqrt{3}\,\mathrm{e}^{\mathrm{i}\frac{11\pi}{8}}$
25.4. $\overline{z}^{\,3}-9z=0$
Passo 1 $\overline{z}^{\,3}=9z$. Com $z=r\,\mathrm{e}^{\mathrm{i}\theta}$: $\;r^{3}\,\mathrm{e}^{-\mathrm{i}\,3\theta}=9r\,\mathrm{e}^{\mathrm{i}\theta}$.
Passo 2 Módulos: $r^{3}=9r\Rightarrow r=0$ ou $r=3$.
Passo 3 Argumentos: $-3\theta=\theta+2k\pi\Leftrightarrow\theta=-\dfrac{k\pi}{2}$.
Os 4 valores distintos: $0,\;-\dfrac{\pi}{2},\;-\pi,\;-\dfrac{3\pi}{2}$.
$z=0\;\vee\;z=3\,\mathrm{e}^{\mathrm{i}\times 0}\;\vee\;z=3\,\mathrm{e}^{-\mathrm{i}\frac{\pi}{2}}\;\vee\;z=3\,\mathrm{e}^{-\mathrm{i}\pi}\;\vee\;z=3\,\mathrm{e}^{-\mathrm{i}\frac{3\pi}{2}}$
25.5. $z^{3}=(2+\mathrm{i})^{3}$
Passo 1 Escrevendo $z=(2+\mathrm{i})\,\zeta$, vem $\zeta^{3}=1$, isto é, $\zeta$ é uma raiz cúbica da unidade.
$\zeta\in\left\{1\,,\,-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\mathrm{i}\,,\,-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\mathrm{i}\right\}$.
Passo 2 Calcular cada produto $(2+\mathrm{i})\,\zeta$:
• $\zeta=1$: $\;z=2+\mathrm{i}$.
• $\zeta=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\mathrm{i}$:
$\;(2+\mathrm{i})\!\left(-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\mathrm{i}\right)=-1+\sqrt{3}\,\mathrm{i}-\dfrac{1}{2}\mathrm{i}-\dfrac{\sqrt{3}}{2}=\left(-1-\dfrac{\sqrt{3}}{2}\right)+\left(\sqrt{3}-\dfrac{1}{2}\right)\mathrm{i}$.
• $\zeta=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\mathrm{i}$:
$\;(2+\mathrm{i})\!\left(-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\mathrm{i}\right)=\left(-1+\dfrac{\sqrt{3}}{2}\right)+\left(-\sqrt{3}-\dfrac{1}{2}\right)\mathrm{i}$.
$z=2+\mathrm{i}\;;\;z=\!\left(-1-\dfrac{\sqrt{3}}{2}\right)+\!\left(\sqrt{3}-\dfrac{1}{2}\right)\mathrm{i}\;;\;z=\!\left(-1+\dfrac{\sqrt{3}}{2}\right)+\!\left(-\sqrt{3}-\dfrac{1}{2}\right)\mathrm{i}$
