How do you simplify $20 (cos (\frac{7 π}{6}) + i sin (\frac{7 π}{6})) \div 15 (cos (\frac{11 π}{3}) + i sin (\frac{11 π}{3}))$ $20(cos((7pi)/6)+isin((7pi)/6))div15(cos((11pi)/3)+isin((11pi)/3))$ and express the result in rectangular form?

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Answer 1 · 2017-01-17T13:09:40

$= - \frac{4}{3} i$ $= - 4/3 i$

In polar complex, that's:

$\frac{20 e^{i \frac{7 π}{6}}}{15 e^{i \frac{11 π}{3}}}$ $(20 e^(i(7 pi)/6))/(15 e^(i (11 pi)/3))$

$= \frac{4}{3} e^{- \frac{5 π}{2}}$ $= 4/3 e^(-(5 pi)/2)$

$= \frac{4}{3} e^{- \frac{π}{2}}$ $= 4/3 e^(-( pi)/2)$

$= - \frac{4 i}{3}$ $= - (4i)/3$

How do you simplify 20(cos((7pi)/6)+isin((7pi)/6))div15(cos((11pi)/3)+isin((11pi)/3))20(cos(7π6)+isin(7π6))÷15(cos(11π3)+isin(11π3))20(cos((7pi)/6)+isin((7pi)/6))div15(cos((11pi)/3)+isin((11pi)/3)) and express the result in rectangular form?