How do you divide $\frac{5 + 2 i}{7 + i}$ $(5+2i)/(7+i)$ in trigonometric form?

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How do you divide $\frac{5 + 2 i}{7 + i}$ $(5+2i)/(7+i)$ in trigonometric form?

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Answer 1 · 2017-04-10T17:17:33

Division of two complex numbers in trigonometric form is defined as:

$\frac{r_{1} (cos (θ_{1}) + i sin (θ_{1}))}{r_{2} (cos (θ_{2}) + i sin (θ_{2}))} = \frac{r_{1}}{r_{2}} (cos (θ_{1} - θ_{2}) + i sin (θ_{1} - θ_{2}))$ $(r_1(cos(theta_1)+isin(theta_1)))/(r_2(cos(theta_2)+isin(theta_2)))= r_1/r_2(cos(theta_1-theta_2)+isin(theta_1-theta_2))$

Explanation:

Given: $\frac{5 + 2 i}{7 + i}$ $(5+2i)/(7+i)$

We need to convert the dividend and the divisor into trigonometric form.

$r_{1} = \sqrt{a_{1}^{2} + b_{1}^{2}}$ $r_1=sqrt(a_1^2+b_1^2)$

$r_{1} = \sqrt{5^{2} + 2^{2}}$ $r_1=sqrt(5^2+2^2)$

$r_{1} = \sqrt{25 + 4}$ $r_1=sqrt(25+4)$

$r_{1} = \sqrt{29}$ $r_1=sqrt(29)$

Because the signs of "a" and "b" are in the 1st quadrant, we use the equation:

$θ_{1} = {tan}^{- 1} (\frac{b_{1}}{a_{1}})$ $theta_1 = tan^-1(b_1/a_1)$

$θ_{1} = {tan}^{- 1} (\frac{2}{5})$ $theta_1=tan^-1(2/5)$

Note: We would have add $π$ $pi$ for the 2nd or 3rd quadrant and $2 π$ $2pi$ for the 4th.

$r_{2} = \sqrt{a_{2}^{2} + b_{2}^{2}}$ $r_2=sqrt(a_2^2+b_2^2)$

$r_{2} = \sqrt{7^{2} + 1^{2}}$ $r_2=sqrt(7^2+1^2)$

$r_{2} = \sqrt{49 + 1}$ $r_2=sqrt(49+1)$

$r_{2} = \sqrt{50}$ $r_2=sqrt(50)$

Because the signs of "a" and "b" are in the 1st quadrant, we use the equation:

$θ_{2} = {tan}^{- 1} (\frac{b_{2}}{a_{2}})$ $theta_2 = tan^-1(b_2/a_2)$

$θ_{2} = {tan}^{- 1} (\frac{1}{7})$ $theta_2=tan^-1(1/7)$

$\frac{5 + 2 i}{7 + i} =$ $(5+2i)/(7+i)=$

$\frac{\sqrt{29} (cos ({tan}^{- 1} (\frac{2}{5})) + i sin ({tan}^{- 1} (\frac{2}{5})))}{\sqrt{50} (cos ({tan}^{- 1} (\frac{1}{7})) + i sin ({tan}^{- 1} (\frac{1}{7})))} =$ $(sqrt(29)(cos(tan^-1(2/5))+isin(tan^-1(2/5))))/(sqrt(50)(cos(tan^-1(1/7))+isin(tan^-1(1/7))))=$

$\sqrt{\frac{29}{50}} (cos ({tan}^{- 1} (\frac{2}{5}) - {tan}^{- 1} (\frac{1}{7})) + i sin ({tan}^{- 1} (\frac{2}{5}) - {tan}^{- 1} (\frac{1}{7})))$ $sqrt(29/50)(cos(tan^-1(2/5)-tan^-1(1/7))+isin(tan^-1(2/5)-tan^-1(1/7)))$

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How do you divide $\frac{5 + 2 i}{7 + i}$ $(5+2i)/(7+i)$ in trigonometric form?

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How do you divide (5+2i)/(7+i) 5+2i7+i (5+2i)/(7+i) in trigonometric form?

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How do you divide $\frac{5 + 2 i}{7 + i}$ $(5+2i)/(7+i)$ in trigonometric form?