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?

Question

17211 views around the world

You can reuse this answer
Creative Commons License

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#