How do you divide # (5+2i)/(7+i) # in trigonometric form?

Question

1457 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T17:17:33

Division of two complex numbers in trigonometric form is defined as:

#(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))#

Given: #(5+2i)/(7+i)#

We need to convert the dividend and the divisor into trigonometric form.

#r_1=sqrt(a_1^2+b_1^2)#

#r_1=sqrt(5^2+2^2)#

#r_1=sqrt(25+4)#

#r_1=sqrt(29)#

Because the signs of "a" and "b" are in the 1st quadrant, we use the equation:

#theta_1 = tan^-1(b_1/a_1)#

#theta_1=tan^-1(2/5)#

Note: We would have add #pi# for the 2nd or 3rd quadrant and #2pi# for the 4th.

#r_2=sqrt(a_2^2+b_2^2)#

#r_2=sqrt(7^2+1^2)#

#r_2=sqrt(49+1)#

#r_2=sqrt(50)#

Because the signs of "a" and "b" are in the 1st quadrant, we use the equation:

#theta_2 = tan^-1(b_2/a_2)#

#theta_2=tan^-1(1/7)#

#(5+2i)/(7+i)=#

#(sqrt(29)(cos(tan^-1(2/5))+isin(tan^-1(2/5))))/(sqrt(50)(cos(tan^-1(1/7))+isin(tan^-1(1/7))))=#

#sqrt(29/50)(cos(tan^-1(2/5)-tan^-1(1/7))+isin(tan^-1(2/5)-tan^-1(1/7)))#