How do you divide #(i+2) / (9i+14)# in trigonometric form?

1 Answer
Apr 19, 2018

#0.134-0.015i#

Explanation:

For a complex number #z=a+bi# it can be represented as #z=r(costheta+isintheta)# where #r=sqrt(a^2+b^2)# and #theta=tan^-1(b/a)#

#(2+i)/(14+9i)=(sqrt(2^2+1^2)(cos(tan^-1(1/2))+isin(tan^-1(1/2))))/(sqrt(14^2+9^2)(cos(tan^-1(9/14))+isin(tan^-1(9/14))))~~(sqrt5(cos(0.46)+isin(0.46)))/(sqrt277(cos(0.57)+isin(0.57)))#

Given #z_1=r_1(costheta_1+isintheta_1)# and #z_2=r_2(costheta_2+isintheta_2)#, #z_1/z_2=r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))#

#z_1/z_2=sqrt5/sqrt277(cos(0.46-0.57)+isin(0.46-0.57))=sqrt1385/277(cos(-0.11)+isin(-0.11))~~sqrt1385/277(0.99-0.11i)~~0.134-0.015i#

Proof:
#(2+i)/(14+9i)*(14-9i)/(14-9i)=(28-4i+9)/(14^2+9^2)=(37-4i)/277~~0.134-0.014i#