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

1 Answer
Mar 28, 2017

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))#