Question #f593a

Consider the 2 complex numbers.

#z_1=r_1(costheta_1+isintheta_1)#

and #z_2=r_2(costheta_2+isintheta_2)#

dividing them gives.

#(z_1)/(z_2)=(r_1(costheta_1+isintheta_1))/(r_2(costheta_2+isintheta_2))#

multiply numerator and denominator by #(costheta_2-isintheta_2)#

#(z_1)/(z_2)=(r_1(costheta_1+isintheta_1)(costheta_2-isintheta_2))/(r_2(costheta_2+isintheta_2)(costheta_2-isintheta_2))#

distribute the brackets.

#=(r_1(costheta_1costheta_2+sintheta_1sintheta_2+i(sintheta_1costheta_2-costheta_1sintheta_2)))/(r_2(cos^2theta_2+sin^2theta_2))#

#=[(r_1)/(r_2)](cos(theta_1-theta_2)+isin(theta_1-theta_2)........ (A)#

(A) is obtained using the #color(blue)"trigonometric identities"#

#color(red)(|bar(ul(color(white)(a/a)color(black)(cos(A-B)=cosAcosB+sinAsinB)color(white)(a/a)|)))" and "#

#color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A-B)=sinAcosB-cosAsinB)color(white)(a/a)|)))#

and of course, the denominator #cos^2theta_2+sin^2theta_2=1#

The result being #color(red)(|bar(ul(color(white)(a/a)color(black)(|z_1/(z_2)|=(r_1)/(r_2)" and " arg((z_1)/(z_2))=theta_1-theta_2)color(white)(a/a)|)))#
#color(blue)"----------------------------------------------------------------"#

Here #r_1=5,theta_1=(5pi)/8" and " r_2=10,theta_2=pi/4#

#rArr(5cis((5pi)/8))/(10cis(pi/4))=1/2cis((5pi)/8-pi/4)=1/2cis((3pi)/8)#