How do you multiply # (7-3i)(5-i) # in trigonometric form?

1 Answer
Mar 8, 2017

#(7-3i)(5-i)=2sqrt377(costheta+isintheta)#, where #theta=tan^(-1)(-11/16)#

Explanation:

Let us write the two complex numbers in polar coordinates and let them be

#z_1=r_1(cosalpha+isinalpha)# and #z_2=r_2(cosbeta+isinbeta)#

Here, if two complex numbers are #a_1+ib_1# and #a_2+ib_2# #r_1=sqrt(a_1^2+b_1^2)#, #r_2=sqrt(a_2^2+b_2^2)# and #alpha=tan^(-1)(b_1/a_1)#, #beta=tan^(-1)(b_2/a_2)#

Their multipication leads us to

#{r_1*r_2}{(cosalpha+isinalpha)*(cosbeta+isinbeta)}# or

#{r_1*r_2}{(cosalphacosbeta+i^2sinalphasinbeta)+i(cosalphasinbeta+cosbetasinalpha))# or

#{r_1*r_2}{(cosalphacosbeta-sinalphasinbeta)+i(cosalphasinbeta+cosbetasinalpha))# or

#(r_1*r_2)*(cos(alpha+beta)+isin(alpha+beta))# or

#z_1*z_2# is given by #(r_1*r_2, (alpha+beta))#

So for multiplication of complex number #z_1# and #z_2# , take new angle as #(alpha+beta)# and modulus os #r_1*r_2# of the modulus of two numbers.

Here #7-3i# can be written as #r_1(cosalpha+isinalpha)# where #r_1=sqrt(7^2+(-3)^2)=sqrt58# and #alpha=tan^(-1)((-3)/7)#

and #5-i# can be written as #r_2(cosbeta+isinbeta)# where #r_2=sqrt(5^2+(-1)^2)=sqrt26# and #beta=tan^(-1)(-1/5)#

and #z_1*z_2=sqrt58*(sqrt26)(costheta+isintheta)#, where #theta=alpha+beta#

Hence, #tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-3)/7+(-1/5))/(1-((-3)/7xx(-1/5)))=(-22/35)/(32/35)=-22/32=-11/16#.

Hence, #(7-3i)(5-i)=sqrt(58xx26)(costheta+isintheta)#

= #2sqrt377(costheta+isintheta)#, where #theta=tan^(-1)(-11/16)#