How do you simplify #6(cos((3pi)/4)+isin((3pi)/4))div2(cos(pi/4)+isin(pi/4))# and express the result in rectangular form?

1 Answer
Feb 21, 2017

Answer:

#z=3i#

Explanation:

#"Given"#

#z_1=r_1(costheta_1+isintheta_1)" and "#

#z_2=r_2(costheta_2+isintheta_2)" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)((z_1)/(z_2)=(r_1)/(r_2)[cos(theta_1-theta_2)+isin(theta_1-theta_2)])color(white)(2/2)|)))#

#rArr(6[cos((3pi)/4)+isin((3pi)/4)])/(2[cos(pi/4)+isin(pi/4)#

#=6/2[cos((3pi)/4-pi/4)+isin((3pi)/4)-pi/4)]#

#=3[cos(pi/2)+isin(pi/2)]#

#=3(0+i)#

#=0+3i#

#rArrz=3i#