How do you perform the operation in trigonometric form #(0.45(cos(310)+isin(310)))(0.6(cos(200)+isin(200)))#?

1 Answer
Jan 5, 2017

#(0.45(cos310^@+isin310^@))(0.6(cos200^@+isin200^@))=0.27(-sqrt3/2+1/2i)#

Explanation:

Two complex numbers in polar form #z_1=r_1(cosalpha+isinalpha)# and #z_2=r_2(cosbeta+isinbeta)# can be multiplied as under,

#z_1*z_2=r_1*r_2[cosalphacosbeta+icosalphasinbeta+isinalphacosbeta+i^2sinalphasinbeta]#

= #r_1*r_2[(cosalphacosbeta-sinalphasinbeta)+i(cosalphasinbeta+sinalphacosbeta)]#

= #r_1*r_2[cos(alpha+beta)+isin(alpha+beta)]#

Hence #(0.45(cos310^@+isin310^@))(0.6(cos200^@+isin200^@))#

= #0.45xx0.6(cos(310^@+200^@)+isin(310^@+200^@))#

= #0.27(cos510^@+isin510^@)#

= #0.27(cos(360^@+150^@)+isin(360^@+150^@))#

= #0.27(cos150^@+isin150^@)#

= #0.27(-sqrt3/2+1/2i)#