How do you divide #( 2i-6) / ( -12i +4 )# in trigonometric form?

1 Answer
Shwetank Mauria
May 19, 2016

#(2i-6)-:(-12i+4)=1/2(cosrho+isinrho)# where #rho=tan^(-1)(-5/3)#

Explanation:

Let us first write #(2i-6)# and #(-12i+4)# in trigonometric form.

#a+ib# can be written in trigonometric form #re^(itheta)=rcostheta+irsintheta=r(costheta+isintheta)#,
where #r=sqrt(a^2+b^2)# and #tantheta=b/a# or #theta=arctan(b/a)#

Hence #2i-6=(-6+2i)=sqrt((-6)^2+2^2)[cosalpha+isinalpha]# or

#sqrt40e^(ialpha)#, where #tanalpha=(-1)/3# and

#-12i+4=(4-12i)=sqrt(4^2+(-12)^2)[cosbeta+isinbeta]# or

#sqrt160e^(ibeta]#, where #tanbeta=(-12)/4=-3#

Hence #(2i-6)-:(-12i+4)=(sqrt40e^(ialpha))/(sqrt160e^(ibeta])=sqrt(1/4)e^(i(alpha-beta))=e^(i(alpha-beta))/2=1/2(cos(alpha-beta)+isin(alpha-beta))#

Now, #tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)#

= #(-1/3+(-3))/(1+(-1/3)*(-3))=(-10/3)/2=-5/3#

Hence #(2i-6)-:(-12i+4)=1/2(cosrho+isinrho)# where #rho=tan^(-1)(-5/3)#

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