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Loser66

  • one year ago

Find real and imaginary part of \((\dfrac{-1-i\sqrt3}{2})^6\) by polar expression. Please, help

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  1. Loser66
    • one year ago
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    oh, I know what is wrong, it should be \(z = |-1| e^{i4\pi/3}\) right?

  2. IrishBoy123
    • one year ago
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    \((\dfrac{-1-i\sqrt3}{2})^6\)

  3. Loser66
    • one year ago
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    Yes|dw:1441922670446:dw|

  4. anonymous
    • one year ago
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    http://www.wolframalpha.com/widgets/view.jsp?id=47e6fd9e600778e368177b1edd70b025

  5. anonymous
    • one year ago
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    idk why i typed this :O

  6. anonymous
    • one year ago
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    and @Loser66 the polar form is the one with cosine and sine not exponential Form

  7. Loser66
    • one year ago
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    But we can find the real part and imaginary part by polar, right?

  8. Loser66
    • one year ago
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    As above, I have \(z = e^{4i\pi/3}\) , then \(z^6 = e^{i8\pi}= cos (8\pi) + isin(8\pi)\) Hence R(z) =1 and Im (z) =0, right?

  9. anonymous
    • one year ago
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    yeah sure \( (x+iy)^n=(r\cos n \theta+i r \sin n \theta) \)

  10. IrishBoy123
    • one year ago
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    \[(\dfrac{-1-i\sqrt3}{2})^6\] \[= (-\frac{1}{2} -i\frac{\sqrt{3}}{2})^6\] \[=(e^{i \frac{4 \pi}{3}})^6\] \[=e^{i 8 \pi} = 1\]

  11. IrishBoy123
    • one year ago
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    find the 6 sixth roots of 1, next that's more frustrating.

  12. IrishBoy123
    • one year ago
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    |dw:1441924027440:dw|

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