How do you divide #( 3i+8) / (i +4 )# in trigonometric form?

1 Answer
Shwetank Mauria
May 21, 2016

#(6i-4)/(-3i-5)=sqrt(73/17)(cosrho+isinrho)# where #rho=tan^(-1)(4/35)#

Explanation:

Let us first write #(3i+8)# and #(i+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 #3i+8=(8+3i)=sqrt(8^2+3^2)[cosalpha+isinalpha]# or

#sqrt73e^(ialpha)#, where #tanalpha=3/8# and

#i+4=(4+i)=sqrt(4^2+1^2)[cosbeta+isinbeta]# or

#sqrt17e^(ibeta]#, where #tanbeta=1/4#

Hence #(3i+8)/(i+4)=(sqrt73e^(ialpha))/(sqrt17e^(ibeta])=sqrt(73/17)e^(i(alpha-beta))=sqrt(73/17)(cos(alpha-beta)+isin(alpha-beta))#

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

= #(3/8-1/4)/(1+3/8*1/4)=(1/8)/(1+3/32)=1/8*32/35=4/35#

Hence #(6i-4)/(-3i-5)=sqrt(73/17)(cosrho+isinrho)# where #rho=tan^(-1)(4/35)#

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