How do you divide (2-4i)/(5+8i) in trigonometric form?

1 Answer
Jul 4, 2016

(2-4i)/(5+8i)=sqrt(20/89)(costheta+isintheta), where theta=tan^(-1)(18/11)

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 2-4i can be written as r_1(cosalpha+isinalpha) where r_1=sqrt(2^2+(-4)^2)=sqrt20 and alpha=tan^(-1)(-4/2)=tan^(-1)(-2)

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

and z_1/z_2=sqrt20/(sqrt89)(costheta+isintheta), where theta=alpha-beta

Hence, tantheta=tan(alpha-beta)=(tanalpha-tanbeta)/(1+tanalphatanbeta)=((-2)-(8/5))/(1+(-2)xx(8/5))=(-18/5)/(-11/5)=18/11.

Hence, (2-4i)/(5+8i)=sqrt(20/89)(costheta+isintheta), where theta=tan^(-1)(18/11)