How do you divide # (1+7i) / (5-3i) # in trigonometric form?

1 Answer
Oct 28, 2016

#(1+7i)/(5-3i)=5/sqrt17(costheta+isintheta)#, where #theta=tan^(-1)(-19/8)#

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 #1+7i# can be written as #r_1(cosalpha+isinalpha)# where #r_1=sqrt(1^2+7^2)=sqrt50# and #alpha=tan^(-1)7#

and #5-3i# can be written as #r_2(cosbeta+isinbeta)# where #r_2=sqrt(5^2+3^2)=sqrt34# and #beta=tan^(-1)(-3/5)#

and #z_1/z_2=sqrt50/(sqrt34)(costheta+isintheta)#, where #theta=alpha-beta#

Hence, #tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=(7-(-3/5))/(1+7xx(-3/5))=(38/5)/(-16/5)=-19/8#.

Hence, #(1+7i)/(5-3i)=sqrt50/sqrt34(costheta+isintheta)#

= #5/sqrt17(costheta+isintheta)#, where #theta=tan^(-1)(-19/8)#