Question

How do you divide `(2i-7) / (-2i-8)` in trigonometric form?

Answer 1

Division in trigonometric form is:
`(r_1(cos(theta_1) + isin(theta_1)))/(r_2(cos(theta_2) + isin(theta_2)))=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))`

Explanation:

Given: `(2i-7)/(-2i-8)`

`r_1=sqrt((-7)^2+2^2)`

`r_1=sqrt53`

To find the value of `theta_1`, we must observe that the real part is negative and the imaginary part is positive; this places the angle in the 2nd quadrant:

`theta_1=pi+tan^-1(2/-7)`

`theta_1=pi-tan^-1(2/7)`

Moving on to `r_2`:

`r_2 = sqrt((-8)^2+(-2)^2)`

`r_2=sqrt(68)`

To find the value of `theta_2`, we must observe that the real part is negative and the imaginary part is negative; this places the angle in the 3nd quadrant:

`theta_2= pi+tan^-1((-2)/-8)`

`theta_2= pi+tan^-1(1/4)`

`(2i-7)/(-2i-8)=sqrt(53/68)(cos(-tan^-1(2/7)-tan^-1(1/4))+isin(-tan^-1(2/7)-tan^-1(1/4)))`