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

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

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Answer 1 · 2018-01-12T15:02:41

$(\sqrt{\frac{65}{2}}) \cdot e^{(arctan (\frac{4}{7}) - \frac{3 π}{4}) i}$ $(sqrt(65/2))*e^((arctan(4/7)-(3pi)/4)i)$

Explanation:

$\frac{7 + 4 i}{- 1 + i} = \frac{z_{1}}{z_{2}}$ $(7+4i)/(-1+i)=(z_1)/(z_2)$

$ρ_{1} = \sqrt{7^{2} + 4^{2}} = \sqrt{49 + 16} = \sqrt{65}$ $rho_1=sqrt(7^2+4^2)=sqrt(49+16)=sqrt(65)$
$θ_{1} = arctan (\frac{4}{7})$ $theta_1=arctan(4/7)$

$ρ_{1} = \sqrt{1^{2} + 1^{2}} = \sqrt{1 + 1} = \sqrt{2}$ $rho_1=sqrt(1^2+1^2)=sqrt(1+1)=sqrt(2)$
$θ_{1} = arctan (\frac{- 1}{1}) = arctan (- 1) = (\frac{3 π}{4}),$ $theta_1=arctan((-1)/1)=arctan(-1)=((3pi)/4),$

$\frac{z_{1}}{z_{2}} = \frac{ρ_{1}}{ρ_{2}} \cdot e^{(θ_{1} - θ_{2} i)}$ $color(blue)((z_1)/(z_2)=(rho_1)/(rho_2)*e^[(theta_1-theta_2i)]$

$\frac{z_{1}}{z_{2}} = \frac{\sqrt{65}}{\sqrt{2}} \cdot e^{arctan (\frac{4}{7} i) - \frac{3 π}{4} i}$ $(z_1)/(z_2)=(sqrt(65))/sqrt(2)*e^(arctan(4/7i)-(3pi)/4i)$

$\frac{z_{1}}{z_{2}} = (\sqrt{\frac{65}{2}}) \cdot e^{(arctan (\frac{4}{7}) - \frac{3 π}{4})} i$ $(z_1)/(z_2)=(sqrt(65/2))*e^((arctan(4/7)-(3pi)/4))i$

$\frac{z_{1}}{z_{2}} \approx ~ 5.701 \cdot e^{- 1.837 i}$ $(z_1)/(z_2)~~~5.701*e^(-1.837i)$

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

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