Question

How do you divide `(i+2) / (9i+14)` in trigonometric form?

Answer 1

`0.134-0.015i`

Explanation:

For a complex number `z=a+bi` it can be represented as `z=r(costheta+isintheta)` where `r=sqrt(a^2+b^2)` and `theta=tan^-1(b/a)`

`(2+i)/(14+9i)=(sqrt(2^2+1^2)(cos(tan^-1(1/2))+isin(tan^-1(1/2))))/(sqrt(14^2+9^2)(cos(tan^-1(9/14))+isin(tan^-1(9/14))))~~(sqrt5(cos(0.46)+isin(0.46)))/(sqrt277(cos(0.57)+isin(0.57)))`

Given `z_1=r_1(costheta_1+isintheta_1)` and `z_2=r_2(costheta_2+isintheta_2)`, `z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))`

`z_1/z_2=sqrt5/sqrt277(cos(0.46-0.57)+isin(0.46-0.57))=sqrt1385/277(cos(-0.11)+isin(-0.11))~~sqrt1385/277(0.99-0.11i)~~0.134-0.015i`

Proof:
`(2+i)/(14+9i)*(14-9i)/(14-9i)=(28-4i+9)/(14^2+9^2)=(37-4i)/277~~0.134-0.014i`