Question

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

Answer 1

`(-1+2i)*(3-i)=sqrt50[cosarctan(-7)+isinarctan(-7)]`

Explanation:

We first covert them into trigonometric form. In this form `a+bi=r(costheta+isintheta)` or `a+bi=r*e^(itheta)`, where `r=sqrt(a^2+b^2)` and `theta=arctan(b/a)`

Hence, `-1+2i=sqrt5e^(ialpha)`, where `alpha=arctan(-2)`

and `3-i=sqrt10e^(ibeta)`, where `beta=arctan(-1/3)`

Hence `(-1+2i)*(3-i)=sqrt50e^(i(alpha+beta))`

As `tan(alpha+beta)=(tanalpha+tanbeta)/(1-tanalpha*tanbeta)` or

`tan(alpha+beta)=((-2)-1/3)/(1-(-2)*(-1/3))=(-7/3)/(1-2/3)=(-7/3)/(1/3)=-7`

Hence `(-1+2i)*(3-i)=sqrt50e^(i(arctan(-7)))`

or `(-1+2i)*(3-i)=sqrt50[cosarctan(-7)+isinarctan(-7)]`