Question

How do you divide `(-3-4i)/(5+2i)` in trigonometric form?

Answer 1

`5/sqrt(29)(cos(0.540)+isin(0.540))~~0.79+0.48i`

Explanation:

`(-3-4i)/(5+2i)=-(3+4i)/(5+2i)`

`z=a+bi` can be written as `z=r(costheta+isintheta)`, where

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

For `z_1=3+4i`:
`r=sqrt(3^2+4^2)=5`
`theta=tan^-1(4/3)=~~0,927`

For `z_2=5+2i`:
`r=sqrt(5^2+2^2)=sqrt29`
`theta=tan^-1(2/5)=~~0.381`

For `z_1/z_2`:
`z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))`

`z_1/z_2=5/sqrt(29)(cos(0.921-0.381)+isin(0.921-0.381))`

`z_1/z_2=5/sqrt(29)(cos(0.540)+isin(0.540))=0.79+0.48i`

Proof:
`-(3+4i)/(5+2i)*(5-2i)/(5-2i)=-(15+20i-6i+8)/(25+4)=(23+14i)/29=0.79+0.48i`