# Let z1 = 2(cos 5pi/6 + i sin 5pi/6) and z2 = 5(cos pi/3 + i sin pi/3), how do you find z1z2?

Nov 25, 2016

${z}_{1} \times {z}_{2} = - 5 \sqrt{3} - 5 i$

#### Explanation:

If we have two complex numbers ${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

${z}_{1} \times {z}_{2} = {r}_{1} {r}_{2} \left(\cos \alpha \cos \beta + i \cos \alpha \sin \beta + i \sin \alpha \cos \beta + {i}^{2} \sin \alpha \sin \beta\right)$

or
z_1xxz_2=r_1r_2((cosalphacosbeta-sinalphasinbeta)+i(cosalphasinbeta+sinalphacosbeta)

or
${z}_{1} \times {z}_{2} = {r}_{1} {r}_{2} \left(\cos \left(\alpha + \beta\right) + i \sin \left(\alpha + \beta\right)\right)$

Here we have ${z}_{1} = 2 \left(\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right)\right)$ and z_2=5(cos(pi/3)+isin(pi/3)

and ${z}_{1} \times {z}_{2} = 2 \times 5 \left(\cos \left(\frac{5 \pi}{6} + \frac{\pi}{3}\right) + i \sin \left(\frac{5 \pi}{6} + \frac{\pi}{3}\right)\right)$

= $10 \left(\cos \left(\frac{7 \pi}{6}\right) + i \sin \left(\frac{7 \pi}{6}\right)\right)$

= $10 \left(\cos \left(\pi + \frac{\pi}{6}\right) + i \sin \left(\pi + \frac{\pi}{6}\right)\right)$

= $10 \left(- \cos \left(\frac{\pi}{6}\right) - i \sin \left(\frac{\pi}{6}\right)\right)$

= $- 10 \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)$

= $- 5 \sqrt{3} - 5 i$