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)