How do you divide (8-2i) / (5-3i) in trigonometric form?

1 Answer
Leland Adriano Alejandro
Jan 27, 2016

sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23)) OR

sqrt2(cos(16.9275^@)+isin(16.9275^@))

Explanation:

Start from the given complex number

(8-2i)/(5-3i)

Convert the numerator

r_1=sqrt(8^2+(-2)^2)=sqrt68

theta_1=tan^-1 ((-2)/8)=tan^-1 ((-1)/4)

then

8-2i=sqrt68[cos(tan^-1 ((-1)/4))+isin(tan^-1 ((-1)/4))]

Convert the denominator

r_2=sqrt(5^2+(-3)^2)=sqrt34

theta_2=tan^-1 ((-3)/5)=tan^-1 ((-3)/5)

then

5-3i=sqrt34[cos(tan^-1 ((-3)/5))+isin(tan^-1 ((-3)/5))]

Let us divide now, from the given with the equivalent

(8-2i)/(5-3i)=(sqrt68[cos(tan^-1 ((-1)/4))+isin(tan^-1 ((-1)/4))])/(sqrt34[cos(tan^-1 ((-3)/5))+isin(tan^-1 ((-3)/5))])

Divide using the following formula

(r_1/r_2)*[cos (theta_1-theta_2)+isin(theta1-theta_2)]

(8-2i)/(5-3i)= sqrt(68/34)[cos(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))+isin(tan^-1 ((-1)/4)-tan^-1 ((-3)/5))]

Take note: that
tan (theta_1-theta_2)=(tan theta_1-tan theta_2) /(1+tan theta_1*tan theta_2)=(-1/4-(-3/5))/(1+(-1/4)*(-3/5))

tan (theta_1-theta_2)=7/23

theta_1-theta_2=tan^-1 (7/23)

and

(8-2i)/(5-3i)=sqrt2(cos(tan^-1 (7/23))+isin(tan^-1 (7/23))

Have a nice day!!!.

Related questions
See all questions in The Trigonometric Form of Complex Numbers
Impact of this question
1759 views around the world
You can reuse this answer
Creative Commons License