How do you multiply (1-2i)(2-3i) in trigonometric form?

1 Answer
Jul 23, 2016

(1-2i)xx(2-3i)=sqrt65(costheta+isintheta), where theta=tan^(-1)(7/4)

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 1-2i can be written as r_1(cosalpha+isinalpha) where r_1=sqrt(1^2+(-2)^2)=sqrt5 and alpha=tan^(-1)(-2/1)=tan^(-1)(-2)

and 2-3i can be written as r_2(cosbeta+isinbeta) where r_2=sqrt(2^2+(-3)^2)=sqrt(4+9)=sqrt13 and beta=tan^(-1)(-3/2)

and z_1*z_2=sqrt5xxsqrt13(costheta+isintheta), where theta=alpha+beta

Hence, as tantheta=tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalphatanbeta)=((-2)+(-3/2))/(1-(-2)xx(-3/2))=(-7/2)/(1-3)=(-7/2)/(-2)=7/4.

and z=sqrt65

Hence, (1-2i)xx(2-3i)=sqrt65(costheta+isintheta), where theta=tan^(-1)(7/4)

Note that both alpha and beta are in fourth quadrant based on signs of sine and cosine functions, hence theta=alpha+beta ought to be between 3pi and 4pi. Hence as tantheta is positive, theta is in third quadrant.