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

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Answer 1 · 2018-01-12T15:02:41

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

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

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

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

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

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

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

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

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