How do you divide $\frac{2 - 7 i}{2 - 8 i}$ $(2-7i)/(2-8i)$ in trigonometric form?

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How do you divide $\frac{2 - 7 i}{2 - 8 i}$ $(2-7i)/(2-8i)$ in trigonometric form?

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Answer 1 · 2016-01-19T02:26:53

Find the polar form $C_{p} = (R, θ)$ $C_p = (R, theta)$
given $C = x + i y$ $C = x +iy$
Then the polar form is $R = \sqrt{x^{2} + y^{2}}$ $R = sqrt(x^2+y^2)$
And $θ = {tan}^{- 1} (\frac{y}{x})$ $theta= tan^-1 (y/x)$

Thus $R_{1} = \sqrt{53}; θ_{1} = {tan}^{- 1} (\frac{7}{2});$ $R_1= sqrt(53); theta_1= tan^-1(7/2);$

careful on the angle use the sign of the imaginary number to get it right (hint it is in the 4th quadrant...

$R_{2} = \sqrt{68}; θ_{2} = {tan}^{- 1} (\frac{8}{2});$ $R_2= sqrt(68); theta_2 = tan^-1(8/2);$

again make sure you have the right angle.

now divide $R = \frac{R_{1}}{R_{2}}$ $R = R_1/R_2$ and the angle is simply the difference of $θ = θ_{1} - θ_{2}$ $theta = theta_1 - theta_2$

Good luck, hope it helped
Yonas

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How do you divide $\frac{2 - 7 i}{2 - 8 i}$ $(2-7i)/(2-8i)$ in trigonometric form?

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