How do you multiply $(3 + 5 i) (6 - 6 i)$ $(3+5i)(6-6i)$ in trigonometric form?

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How do you multiply $(3 + 5 i) (6 - 6 i)$ $(3+5i)(6-6i)$ in trigonometric form?

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Answer 1 · 2018-07-27T17:23:36

$6 (8 + 2 i)$ $6(8+2i)$

Explanation:

Given that

$(3 + 5 i) (6 - 6 i)$ $(3+5i)(6-6i)$

$= \sqrt{34} (cos ({tan}^{- 1} (\frac{5}{3})) + i sin ({tan}^{- 1} (\frac{5}{3}))) \cdot 6 \sqrt{2} (cos (- \frac{π}{4}) + i sin (- \frac{π}{4}))$ $=\sqrt34(\cos(\tan^{-1}(5/3))+i\sin(\tan^{-1}(5/3)))\cdot 6\sqrt2(\cos(-\pi/4)+i\sin(-\pi/4))$

$= 12 \sqrt{17} (cos ({tan}^{- 1} (\frac{5}{3}) - \frac{π}{4}) + i sin ({tan}^{- 1} (\frac{5}{3}) - \frac{π}{4}))$ $=12\sqrt17(\cos(\tan^{-1}(5/3)-\pi/4)+i\sin(\tan^{-1}(5/3)-\pi/4))$

$= 12 \sqrt{17} (cos ({tan}^{- 1} (\frac{5}{3})) cos (\frac{π}{4}) + sin ({tan}^{- 1} (\frac{5}{3})) sin (\frac{π}{4}) + i (sin ({tan}^{- 1} (\frac{5}{3})) cos (\frac{π}{4}) - cos ({tan}^{- 1} (\frac{5}{3})) sin (\frac{π}{4})))$ $=12\sqrt17(\cos(\tan^{-1}(5/3))\cos(\pi/4)+\sin(\tan^{-1}(5/3))\sin(\pi/4)+i(\sin(\tan^{-1}(5/3))\cos(\pi/4)-\cos(\tan^{-1}(5/3))\sin(\pi/4)))$

$= 12 \sqrt{17} (\frac{3}{\sqrt{34}} \frac{1}{\sqrt{2}} + \frac{5}{\sqrt{34}} \frac{1}{\sqrt{2}} + i (\frac{5}{\sqrt{34}} \frac{1}{\sqrt{2}} - \frac{3}{\sqrt{34}} \frac{1}{\sqrt{2}}))$ $=12\sqrt17(\frac{3}{\sqrt34}1/\sqrt2+5/\sqrt34 1/\sqrt2+i(\frac{5}{\sqrt34}1/\sqrt2-3/\sqrt34 1/\sqrt2))$

$= \frac{12 \sqrt{17}}{2 \sqrt{17}} (8 + 2 i)$ $={12\sqrt17}/{2\sqrt17}(8+2i)$

$= 6 (8 + 2 i)$ $=6(8+2i)$

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How do you multiply $(3 + 5 i) (6 - 6 i)$ $(3+5i)(6-6i)$ in trigonometric form?

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Explanation:

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Math

Humanities

... and beyond

How do you multiply (3+5i)(6−6i) (3+5i)(6-6i) in trigonometric form?

1 Answer

Explanation:

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How do you multiply $(3 + 5 i) (6 - 6 i)$ $(3+5i)(6-6i)$ in trigonometric form?