How do you multiply # (1-2i)(2-3i) # in trigonometric form?

1 Answer
Jul 23, 2016

#(1-2i)xx(2-3i)=sqrt65(costheta+isintheta)#, where #theta=tan^(-1)(7/4)#

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 #1-2i# can be written as #r_1(cosalpha+isinalpha)# where #r_1=sqrt(1^2+(-2)^2)=sqrt5# and #alpha=tan^(-1)(-2/1)=tan^(-1)(-2)#

and #2-3i# can be written as #r_2(cosbeta+isinbeta)# where #r_2=sqrt(2^2+(-3)^2)=sqrt(4+9)=sqrt13# and #beta=tan^(-1)(-3/2)#

and #z_1*z_2=sqrt5xxsqrt13(costheta+isintheta)#, where #theta=alpha+beta#

Hence, as #tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-2)+(-3/2))/(1-(-2)xx(-3/2))=(-7/2)/(1-3)=(-7/2)/(-2)=7/4#.

and #z=sqrt65#

Hence, #(1-2i)xx(2-3i)=sqrt65(costheta+isintheta)#, where #theta=tan^(-1)(7/4)#

Note that both #alpha# and #beta# are in fourth quadrant based on signs of sine and cosine functions, hence #theta=alpha+beta# ought to be between #3pi# and #4pi#. Hence as #tantheta# is positive, #theta# is in third quadrant.