How do you multiply # (-1+8i)(-9+7i) # in trigonometric form?

1 Answer
Jul 23, 2016

#(-1+8i)xx(-9+7i)=65sqrt2(costheta+isintheta)#, where #theta=tan^(-1)(79/47)#

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 multiplicaton leads us to

#{r_1xxr_2}{(cosalpha+isinalpha)xx(cosbeta+isinbeta)}# or

#{r_1r_2}(cosalphacosbeta+isinalphacosbeta+isinalphacosbeta+i^2sinalphasinbeta)#

#{r_1r_2}(cosalphacosbeta+isinalphacosbeta+isinalphacosbeta-sinalphasinbeta)#

#{r_1r_2}[(cosalphacosbeta-sinalphasinbeta+i(sinalphacosbeta+sinalphacosbeta)]# or

#(r_1r_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 #r_1*r_2# of the modulus of two numbers.

Here #-1+8i# can be written as #r_1(cosalpha+isinalpha)# where #r_1=sqrt((-1)^2+8^2)=sqrt65# and #alpha=tan^(-1)(-8/1)=tan^(-1)(-8)#

and #-9+7i# can be written as #r_2(cosbeta+isinbeta)# where #r_2=sqrt((-9)^2+7^2)=sqrt(81+49)=sqrt130# and #beta=tan^(-1)(7/(-9))=tan^(-1)(-7/9)#

and #z_1*z_2=sqrt65xxsqrt130(costheta+isintheta)#, where #theta=alpha+beta#

Hence, as #tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-8)+(-7/9))/(1-(-8)xx(-7/9))=(-79/9)/(1-(56/9))=(-79/9)/(-47/9)=79/47#.

and #z=65sqrt2#

Hence, #(-1+8i)xx(-9+7i)=65sqrt2(costheta+isintheta)#, where #theta=tan^(-1)(79/47)#