How do you multiply $(- 1 - 6 i) (4 - i)$ $(-1-6i)(4-i)$ in trigonometric form?

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How do you multiply $(- 1 - 6 i) (4 - i)$ $(-1-6i)(4-i)$ in trigonometric form?

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Answer 1 · 2017-03-28T19:12:24

Multiplication of two numbers in trigonometric form:

$r_{1} (cos (θ_{1}) + i sin (θ_{1})) \times r_{2} (cos (θ_{2}) + i sin (θ_{2})) = r_{1} r_{2} (cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$ $r_1(cos(theta_1)+isin(theta_1))xxr_2(cos(theta_2)+isin(theta_2)) = r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))$

Explanation:

For the given numbers:

$r_{1} = \sqrt{- 1^{2} + - 6^{2}}$ $r_1 = sqrt(-1^2+ -6^2)$

$r_{1} = \sqrt{37}$ $r_1 = sqrt(37)$

$θ_{1} = {tan}^{- 1} (\frac{- 6}{- 1}) + π$ $theta_1 = tan^-1((-6)/-1)+pi$

$θ_{1} = {tan}^{- 1} (6) + π$ $theta_1 = tan^-1(6)+pi$

Please notice that we add $π$ $pi$ because the signs of "a" and "b" tell us that the angle is in the 3rd quadrant.

$r_{2} = \sqrt{4^{2} + - 1^{2}}$ $r_2 = sqrt(4^2 + -1^2)$

$r_{2} = \sqrt{17}$ $r_2 = sqrt(17)$

$θ_{2} = {tan}^{- 1} (\frac{- 1}{4}) + 2 π$ $theta_2 = tan^-1((-1)/4)+2pi$

Please notice that we add $2 π$ $2pi$ because the signs of "a" and "b" tell us that the angle is in the 4th quadrant.

Do the multiplication:

$r_{1} (cos (θ_{1}) + i sin (θ_{1})) \times \sqrt{17} (cos (θ_{2}) + i sin (θ_{2})) = r_{1} r_{2} (cos (θ_{1} + θ_{2}) + i sin (θ_{1} + θ_{2}))$ $r_1(cos(theta_1)+isin(theta_1))xxsqrt(17)(cos(theta_2)+isin(theta_2)) = r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))$

$sqrt(37)(cos(tan^-1(6)+pi)+isin(tan^-1(6)+pi))xxsqrt(17)(cos(tan^-1((-1)/4)+2pi)+isin(tan^-1((-1)/4)+2pi)) = sqrt(37)sqrt(17)(cos(tan^-1(6)+pi+tan^-1((-1)/4)+2pi)+isin(tan^-1(6)+pi+tan^-1((-1)/4)+2pi))$

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How do you multiply $(- 1 - 6 i) (4 - i)$ $(-1-6i)(4-i)$ in trigonometric form?

1 Answer

Explanation:

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How do you multiply (-1-6i)(4-i) (−1−6i)(4−i) (-1-6i)(4-i) in trigonometric form?

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How do you multiply $(- 1 - 6 i) (4 - i)$ $(-1-6i)(4-i)$ in trigonometric form?