Question

How do you multiply `(-1-6i)(4-i)` in trigonometric form?

Answer 1

Multiplication of two numbers in trigonometric form:

`r_1(cos(theta_1)+isin(theta_1))xxr_2(cos(theta_2)+isin(theta_2)) = r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))`

Explanation:

For the given numbers:

`r_1 = sqrt(-1^2+ -6^2)`

`r_1 = sqrt(37)`

`theta_1 = tan^-1((-6)/-1)+pi`

`theta_1 = tan^-1(6)+pi`

Please notice that we add `pi` because the signs of "a" and "b" tell us that the angle is in the 3rd quadrant.

`r_2 = sqrt(4^2 + -1^2)`

`r_2 = sqrt(17)`

`theta_2 = tan^-1((-1)/4)+2pi`

Please notice that we add `2pi` because the signs of "a" and "b" tell us that the angle is in the 4th quadrant.

Do the multiplication:

`r_1(cos(theta_1)+isin(theta_1))xxsqrt(17)(cos(theta_2)+isin(theta_2)) = r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))`

`sqrt(37)(cos(tan^-1(6)+pi)+isin(tan^-1(6)+pi))xxsqrt(17)(cos(tan^-1((-1)/4)+2pi)+isin(tan^-1((-1)/4)+2pi)) = sqrt(37)sqrt(17)(cos(tan^-1(6)+pi+tan^-1((-1)/4)+2pi)+isin(tan^-1(6)+pi+tan^-1((-1)/4)+2pi))`