How do you divide #( 2i-1) / ( 3i +5 )# in trigonometric form?

1 Answer
Shwetank Mauria
May 21, 2016

#(2i-1)/(3i+5)=sqrt(5/34)(cosrho+isinrho)# where #rho=tan^(-1)13#

Explanation:

Let us first write #(2i-1)# and #(3i+5)# 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-1=(-1+2i)=sqrt((-1)^2+2^2)[cosalpha+isinalpha]# or

#sqrt5e^(ialpha)#, where #tanalpha=6/(-4)=2/(-1)=-2# and

#3i+5=(5+3i)=sqrt(5^2+3^2)[cosbeta+isinbeta]# or

#sqrt34e^(ibeta]#, where #tanbeta=3/5#

Hence #(2i-1)/(3i+5)=(sqrt5e^(ialpha))/(sqrt34e^(ibeta])=sqrt(5/34)e^(i(alpha-beta))=sqrt(5/34)(cos(alpha-beta)+isin(alpha-beta))#

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

= #(-2-3/5)/(1+(-2)*(3/5))# = #(-13/5)/(-1/5)=-13/5*(-5)/1=13#

Hence #(2i-1)/(3i+5)=sqrt(5/34)(cosrho+isinrho)# where #rho=tan^(-1)13#

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