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?

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Answer 1 · 2016-11-25T05:39:53

$z_1xxz_2=-5sqrt3-5i$

If we have two complex numbers $z_1=r_1(cosalpha+isinalpha)$ and $z_2=r_2(cosbeta+isinbeta)$

$z_1xxz_2=r_1r_2(cosalphacosbeta+icosalphasinbeta+isinalphacosbeta+i^2sinalphasinbeta)$

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

or
$z_1xxz_2=r_1r_2(cos(alpha+beta)+isin(alpha+beta))$

Here we have $z_1=2(cos((5pi)/6)+isin((5pi)/6))$ and $z_2=5(cos(pi/3)+isin(pi/3)$

and $z_1xxz_2=2xx5(cos((5pi)/6+pi/3)+isin((5pi)/6+pi/3))$

= $10(cos((7pi)/6)+isin((7pi)/6))$

= $10(cos(pi+pi/6)+isin(pi+pi/6))$

= $10(-cos(pi/6)-isin(pi/6))$

= $-10(sqrt3/2+i/2)$

= $-5sqrt3-5i$