How do you multiply (-1+8i)(-9+7i) in trigonometric form?

1 Answer
Jul 23, 2016

(-1+8i)xx(-9+7i)=65sqrt2(costheta+isintheta), where theta=tan^(-1)(79/47)

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 multiplicaton leads us to

{r_1xxr_2}{(cosalpha+isinalpha)xx(cosbeta+isinbeta)} or

{r_1r_2}(cosalphacosbeta+isinalphacosbeta+isinalphacosbeta+i^2sinalphasinbeta)

{r_1r_2}(cosalphacosbeta+isinalphacosbeta+isinalphacosbeta-sinalphasinbeta)

{r_1r_2}[(cosalphacosbeta-sinalphasinbeta+i(sinalphacosbeta+sinalphacosbeta)] or

(r_1r_2)(cos(alpha+beta)+isin(alpha+beta)) or

z_1*z_2 is given by (r_1*r_2, (alpha+beta))

So for multiplication of complex number z_1 and z_2 , take new angle as (alpha+beta) and modulus r_1*r_2 of the modulus of two numbers.

Here -1+8i can be written as r_1(cosalpha+isinalpha) where r_1=sqrt((-1)^2+8^2)=sqrt65 and alpha=tan^(-1)(-8/1)=tan^(-1)(-8)

and -9+7i can be written as r_2(cosbeta+isinbeta) where r_2=sqrt((-9)^2+7^2)=sqrt(81+49)=sqrt130 and beta=tan^(-1)(7/(-9))=tan^(-1)(-7/9)

and z_1*z_2=sqrt65xxsqrt130(costheta+isintheta), where theta=alpha+beta

Hence, as tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-8)+(-7/9))/(1-(-8)xx(-7/9))=(-79/9)/(1-(56/9))=(-79/9)/(-47/9)=79/47.

and z=65sqrt2

Hence, (-1+8i)xx(-9+7i)=65sqrt2(costheta+isintheta), where theta=tan^(-1)(79/47)