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Forma trigonométrica do quociente:

Se ${z_1} = {\rho _1}\operatorname{cis} \left( {{\theta _1}} \right)$ e ${z_2} = {\rho _2}\operatorname{cis} \left( {{\theta _2}} \right)$ são dois números complexos não nulos, então $$\frac{{{z_1}}}{{{z_2}}} = \frac{{{\rho _1}}}{{{\rho _2}}}\operatorname{cis} \left( {{\theta _1} – {\theta _2}} \right)$$

1. Ora,
   $$\begin{array}{*{20}{l}}
   {\frac{{2\operatorname{cis} \frac{\pi }{3}}}{{4\operatorname{cis} \left( {\frac{{2\pi }}{3}} \right)}}}& = &{\frac{2}{4}\operatorname{cis} \left( {\frac{\pi }{3} – \frac{{2\pi }}{3}} \right)} \\
   {}& = &{\frac{1}{2}\operatorname{cis} \left( { – \frac{\pi }{3}} \right)} \\
   {}& = &{\frac{1}{2}\operatorname{cis} \frac{{5\pi }}{3}}
   \end{array}$$
2. Ora,
   $$\begin{array}{*{20}{l}}
   {\frac{{ – 2}}{{\operatorname{cis} \left( { – \theta } \right)}}}& = &{\frac{{2\operatorname{cis} \pi }}{{\operatorname{cis} \left( { – \theta } \right)}}} \\
   {}& = &{\frac{2}{1}\operatorname{cis} \left( {\pi  + \theta } \right)} \\
   {}& = &{2\operatorname{cis} \left( {\pi  + \theta } \right)}
   \end{array}$$
3. Ora,
   $$\begin{array}{*{20}{l}}
   {\frac{{ – \operatorname{cis} \frac{\pi }{6}}}{{2\operatorname{cis} \theta }}}& = &{\frac{{\operatorname{cis} \left( {\pi  + \frac{\pi }{6}} \right)}}{{2\operatorname{cis} \theta }}} \\
   {}& = &{\frac{1}{2}\operatorname{cis} \left( {\pi  + \frac{\pi }{6} – \theta } \right)} \\
   {}& = &{\frac{1}{2}\operatorname{cis} \left( {\frac{{7\pi }}{6} – \theta } \right)}
   \end{array}$$
4. Ora,
   $$\begin{array}{*{20}{l}}
   {\left( {2\operatorname{cis} \frac{{5\pi }}{6}} \right) \times \left[ {3\operatorname{cis} \left( { – \frac{{2\pi }}{3}} \right)} \right]}& = &{\left( {2 \times 3} \right)\operatorname{cis} \left( {\frac{{5\pi }}{6} + \left( { – \frac{{2\pi }}{3}} \right)} \right)} \\
   {}& = &{6\operatorname{cis} \frac{\pi }{6}}
   \end{array}$$