How do you divide #( 2i -7) / (- 5 i + 8 )# in trigonometric form?

1 Answer
Dec 8, 2016

#(2i-7)/(-5i+8)=sqrt53/sqrt89(costheta+isintheta)#

= #0.7717(costheta+isintheta)#, where #theta=tan^(-1)(19/66)#

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 division leads us to

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

#{r_1/r_2}{(cosalpha+isinalpha)/(cosbeta+isinbeta)xx(cosbeta-isinbeta)/(cosbeta-isinbeta)}#

#(r_1/r_2){(cosalphacosbeta+sinalphasinbeta)+i(sinalphacosbeta-cosalphasinbeta))/((cos^2beta+sin^2beta))# 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 division complex number #z_1# by #z_2# , take new angle as #(alpha-beta)# and modulus the ratio #r_1/r_2# of the modulus of two numbers.

Here #2i-7# or #-7+2i# can be written as #r_1(cosalpha+isinalpha)# where #r_1=sqrt((-7)^2+2^2)=sqrt53# and #alpha=tan^(-1)(-2/7)#

and #-5i+8# or #8-5i# can be written as #r_2(cosbeta+isinbeta)# where #r_2=sqrt(5^2+8^2)=sqrt89# and #beta=tan^(-1)(-5/8)#

and #z_1/z_2=sqrt53/(sqrt89)(costheta+isintheta)#, where #theta=alpha-beta#

Hence, #tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=((-2/7)-(-5/8))/(1+(-2/7)xx(-5/8))=(19/56)/(66/56)=-19/66#.

Hence, #(2i-7)/(-5i+8)=sqrt53/sqrt89(costheta+isintheta)#

= #0.7717(costheta+isintheta)#, where #theta=tan^(-1)(19/66)#