Question

How do you multiply `(4+5i)(-3+7i)` in trigonometric form?

Answer 1

`(4+5i)(-3+7i)=48.765cis2.317`
`(4+5i)(-3+7i)=-33.111+35.842i`

Explanation:

Let
`z_1=4+5i`
`z_2=-3+7i`
We need to find

`z=z_1z_2`
`r_1=sqrt(4^2+5^2)=sqrt(16+25)=sqrt41`
`r1=sqrt41`
`theta_1=tan^-1(5/4)`
`z_1=r_1cistheta_1`
`z_1=sqrt41cis(tan^-1(5/4))`

`r_2=sqrt((-3)^2+7^2)=sqrt(9+49)=sqrt58`
`r_2=sqrt58`
`theta_2=tan^-1(7/-3)`
`z_2=r_2cistheta_2`
`z_2=sqrt58cis(tan^-1(7/-3))`

`z=z_1z_2`
`z=(sqrt41cis(tan^-1(5/4)))(sqrt58cis(tan^-1(7/-3)))`

`z=sqrt41sqrt58cis(tan^-1(5/4))cis(tan^-1(7/-3))`
`sqrt41sqrt58=sqrt(41xx58)=sqrt2378`
By De-Moivre's Theorem

`cis(tan^-1(5/4))cis(tan^-1(7/-3))=cis(tan^-1(5/4)+tan^-1(7/-3))`
`tan^-1(5/4)+tan^-1(7/-3)=tan^-1((5/4+7/-3)/(1-5/4xx7/-3))`

`5/4+7/-3=(5xx-3+7xx4)/(4xx-3)=(-15+28)/-12=13/-12`
`z=rcistheta`

`r=sqrt2378`
`theta=tan^-1(13/-12)`
`z=sqrt2378cis(tan^-1(13/-12))`
`r=48.765`
`tan^-1(13/-12)`
adjacent side is negative,
the angle lies in the second quadrant
`tan^-1(13/-12)=pi-tan^-1(13/12)`
`tan^-1(13/12)=0.825`
expressed in radians
`tan^-1(13/12)=pi-0.825`
`theta=2.317`

`cos0.825=-0.679`
since the complex number is in 2nd quadrant
`sin0.825=0.735`

`r=48.765`
`x=rcostheta=48.765xx(-0.679)`
`x=-33.111`

`y=rsintheta=48.765xx0.735`
`y=35.842`
`z=x+iy`

`z=(-33.111)+i(35.842)`

Thus,
`(4+5i)(-3+7i)=48.765cis2.317`
`(4+5i)(-3+7i)=-33.111+35.842i`