Question

How do you add `(7+5i)+(2-4i)` in trigonometric form?

Answer 1

`sqrt74cos(0.6202)+sqrt20cos(5.1760)+i(sqrt74sin(0.6202)+sqrt20sin(5.1760))`
`9+i`

Explanation:

`z=a+bi=r(costheta+isintheta)`

`r=sqrt(a^2+b^2)`
`theta=tan^-1(b/a)`

`r_1(cos(theta_1)+isin(theta_2))+r_2(cos(theta_2)+isin(theta_2))=r_1cos(theta_1)+r_2cos(theta_2)+i(r_1sin(theta_1)+r_2sin(theta_2))`

`r_1=sqrt(7^2+5^2)=sqrt(49+25)=sqrt74`
`r_2=sqrt(2^2+4^2)=sqrt(4+16)=sqrt20`

`theta_1=tan^-1(5/7)~~0.6202^c`
`theta_2=tan^-1(-2)~~-1.1071^c`

However, as `2-4i` is in quadrant 4, we need to add `2pi` to get a positive angle variant.

`theta_2=2pi+tan^-1(-2)~~5.1760^c`

`sqrt74cos(0.6202)+sqrt20cos(5.1760)+i(sqrt74sin(0.6202)+sqrt20sin(5.1760))`
`7+2+i(5-4)`
`9+i`

Proof:

`7+5i+2-4`
`(7+2)+i(5-4)`
`9+i`