How do you divide # (7+4i)/(-1+i) # in trigonometric form?

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Answer 1 · 2018-01-12T15:02:41

#(sqrt(65/2))*e^((arctan(4/7)-(3pi)/4)i)#

#(7+4i)/(-1+i)=(z_1)/(z_2)#

#rho_1=sqrt(7^2+4^2)=sqrt(49+16)=sqrt(65)#
#theta_1=arctan(4/7)#

#rho_1=sqrt(1^2+1^2)=sqrt(1+1)=sqrt(2)#
#theta_1=arctan((-1)/1)=arctan(-1)=((3pi)/4),#

#color(blue)((z_1)/(z_2)=(rho_1)/(rho_2)*e^[(theta_1-theta_2i)]#

#(z_1)/(z_2)=(sqrt(65))/sqrt(2)*e^(arctan(4/7i)-(3pi)/4i)#

#(z_1)/(z_2)=(sqrt(65/2))*e^((arctan(4/7)-(3pi)/4))i#

#(z_1)/(z_2)~~~5.701*e^(-1.837i)#