Considere os números na forma trigonométrica:
z_{1}=2 \operatorname{cis}\left(\frac{\pi}{4}\right), \quad z_{2}=3 \operatorname{cis}\left(\frac{\pi}{6}\right)
(a) Calcule z_{1} \cdot z_{2} na forma trigonométrica.
(b) Escreva o resultado também na forma algébrica.
(c) Calcule \frac{z_{1}}{z_{2}} na forma trigonométrica.
(d) Escreva o resultado também na forma algébrica.

Passo 1

O enunciado pede para multiplicar e dividir dois números complexos dados na forma trigonométrica
z_{1}=2\operatorname{cis}\Bigl(\tfrac{\pi}{4}\Bigr), \quad z_{2}=3\operatorname{cis}\Bigl(\tfrac{\pi}{6}\Bigr)
Usaremos as propriedades
  1. Produto: \;|z_{1}z_{2}|=|z_{1}|\cdot|z_{2}|, \arg(z_{1}z_{2})=\arg(z_{1})+\arg(z_{2}).
  2. Quociente: \;\bigl|\tfrac{z_{1}}{z_{2}}\bigr|=\tfrac{|z_{1}|}{|z_{2}|}, \arg\bigl(\tfrac{z_{1}}{z_{2}}\bigr)=\arg(z_{1})-\arg(z_{2}).

Passo 2

Calcular z_{1}\cdot z_{2} na forma trigonométrica.
|z_{1}z_{2}|=2\cdot3=6, \quad \arg(z_{1}z_{2})=\frac{\pi}{4}+\frac{\pi}{6} =\frac{3\pi+2\pi}{12} =\frac{5\pi}{12}.
Logo
z_{1}\cdot z_{2} =6\,\operatorname{cis}\!\Bigl(\tfrac{5\pi}{12}\Bigr).

Passo 3

Converter z_{1}\cdot z_{2} para a forma algébrica.
Usamos
\cos\!\Bigl(\tfrac{5\pi}{12}\Bigr) =\frac{\sqrt6-\sqrt2}{4}, \quad \sin\!\Bigl(\tfrac{5\pi}{12}\Bigr) =\frac{\sqrt6+\sqrt2}{4}.
Então
z_{1}\cdot z_{2} =6\Bigl[\cos\!\bigl(\tfrac{5\pi}{12}\bigr) + i\sin\!\bigl(\tfrac{5\pi}{12}\bigr)\Bigr] =6\Bigl[\frac{\sqrt6-\sqrt2}{4} + i\,\frac{\sqrt6+\sqrt2}{4}\Bigr] =\frac{3}{2}(\sqrt6-\sqrt2) + i\,\frac{3}{2}(\sqrt6+\sqrt2).

Passo 4

Calcular \frac{z_{1}}{z_{2}} na forma trigonométrica.
\bigl|\tfrac{z_{1}}{z_{2}}\bigr| =\frac{2}{3}, \quad \arg\!\bigl(\tfrac{z_{1}}{z_{2}}\bigr) =\frac{\pi}{4}-\frac{\pi}{6} =\frac{3\pi-2\pi}{12} =\frac{\pi}{12}.
Logo
\frac{z_{1}}{z_{2}} =\frac{2}{3}\,\operatorname{cis}\!\Bigl(\tfrac{\pi}{12}\Bigr).

Passo 5

Converter \frac{z_{1}}{z_{2}} para a forma algébrica.
Usamos
\cos\!\Bigl(\tfrac{\pi}{12}\Bigr) =\frac{\sqrt6+\sqrt2}{4}, \quad \sin\!\Bigl(\tfrac{\pi}{12}\Bigr) =\frac{\sqrt6-\sqrt2}{4}.
Então
\frac{z_{1}}{z_{2}} =\frac{2}{3}\Bigl[\frac{\sqrt6+\sqrt2}{4} + i\,\frac{\sqrt6-\sqrt2}{4}\Bigr] =\frac{\sqrt6+\sqrt2}{6} + i\,\frac{\sqrt6-\sqrt2}{6}.

Resposta

(a) z_{1}\cdot z_{2}=6\,\operatorname{cis}\!\bigl(\tfrac{5\pi}{12}\bigr).
(b) z_{1}\cdot z_{2}=\tfrac{3}{2}(\sqrt6-\sqrt2)+i\,\tfrac{3}{2}(\sqrt6+\sqrt2).
(c) \tfrac{z_{1}}{z_{2}}=\tfrac{2}{3}\,\operatorname{cis}\!\bigl(\tfrac{\pi}{12}\bigr).
(d) \tfrac{z_{1}}{z_{2}}=\tfrac{\sqrt6+\sqrt2}{6}+i\,\tfrac{\sqrt6-\sqrt2}{6}.

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