Question

Sendo Z1 = 3•(cos20°+i sen20°) e Z2 = 2•(cos25°+i sen25°), então quanto vale Z1•Z2

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo-a-passo:

$\sf z_1 = 3(cos\:20\textdegree + i\:sen\:20\textdegree)$

$\sf z_2 = 2(cos\:25\textdegree + i\:sen\:25\textdegree)$

$\sf z_1.z_2 = 3 \times 2(cos\:(20\textdegree + 25\textdegree) + i\:sen\:(20\textdegree + 25\textdegree))$

$\boxed{\boxed{\sf z_1.z_2 = 6(cos\:45\textdegree + i\:sen\:45\textdegree)}}$

Answer 2

Explicação passo-a-passo:

De modo geral:

$\sf z_1\cdot z_2=|z_1|\cdot|z_2|\cdot[cos(\theta_{1}+\theta_{2})+i\cdot sen(\theta_{1}+\theta_{2})]$

Assim:

$\sf z_1\cdot z_2=|3|\cdot|2|\cdot[cos(20^{\circ}+25^{\circ})+i\cdot sen(20^{\circ}+25^{\circ})]$

$\sf \red{z_1\cdot z_2=6\cdot(cos~45^{\circ}+i\cdot sen~45^{\circ})}$

Na forma algébrica:

$\sf z_1\cdot z_2=6\cdot\Big(\dfrac{\sqrt{2}}{2}+i\cdot\dfrac{\sqrt{2}}{2}\Big)$

$\sf z_1\cdot z_2=\dfrac{6\sqrt{2}}{2}+i\cdot\dfrac{6\sqrt{2}}{2}$

$\sf \red{z_1\cdot z_2=3\sqrt{2}+3\sqrt{2}\cdot i}$