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anonymous

  • 4 years ago

using Demoivre's theorem solve this equation .. (2+2i)^6

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  1. amistre64
    • 4 years ago
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    thats the z(cos sin ) stuff right?

  2. amistre64
    • 4 years ago
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    -512i if your lucky :)

  3. amistre64
    • 4 years ago
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    using ( ) as delimiters instead of [ ] places it inline

  4. amistre64
    • 4 years ago
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    if we type it out like this \(\frac{a}{b}\) it should go inline text

  5. anonymous
    • 4 years ago
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    thank you but how can i work it out?

  6. 2bornot2b
    • 4 years ago
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    work what out?

  7. anonymous
    • 4 years ago
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    i dont understand... doesnt the 2^6 have to be a square root?

  8. amistre64
    • 4 years ago
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    2^6 = 1 2 4 8 16 32 64.... 64 cos(6pi) = 1 and sin(6pi)= 0 how the wolf gets -516i from that i cant tell yet

  9. amistre64
    • 4 years ago
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    isnt cos(2pi) = 1?

  10. amistre64
    • 4 years ago
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    1+i = pi/4 i believe is a better item

  11. amistre64
    • 4 years ago
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    1 over 1i up is a 45 tri

  12. amistre64
    • 4 years ago
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    or i might be off in my thinking there

  13. amistre64
    • 4 years ago
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    \[(2+2i)^6=2^6(\cos(\pi/4)+i\sin(\pi/4))^6=2^6(\cos(6\pi/4)+i\sin(6\pi/4))\] 64(\cos(3\pi/2)+i\sin(3\pi/2))\] \[64(0-i)\] ugh

  14. amistre64
    • 4 years ago
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    i got the 270^o yay!!

  15. amistre64
    • 4 years ago
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    apparently the radius aint 2

  16. anonymous
    • 4 years ago
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    but how do you get -512i from 64(0-i)?

  17. amistre64
    • 4 years ago
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    the 2^6 aint correct; the (0-i) parts good

  18. amistre64
    • 4 years ago
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    or at least thats what im having an issue with

  19. amistre64
    • 4 years ago
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    2^9 = 512

  20. amistre64
    • 4 years ago
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    that still gives a radius of 64 and not 512 tho

  21. EarthCitizen
    • 4 years ago
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    \[(\sqrt{8})^6 <(45*6 )\]

  22. anonymous
    • 4 years ago
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    where do you get sqrt 8?

  23. amistre64
    • 4 years ago
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    the original radius is 2sqrt(2); not 2 [2sqrt(2)]^6 = 64*8 = 512

  24. amistre64
    • 4 years ago
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    we have to power up the origonal radius

  25. amistre64
    • 4 years ago
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    i knew if i throew demovers out the window I could think of it lol

  26. amistre64
    • 4 years ago
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    \[r^n(cos(nx)+isin(nx))\]

  27. EarthCitizen
    • 4 years ago
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    \[r ^{n}<n \theta\]

  28. 2bornot2b
    • 4 years ago
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    \[(2+2i)=2\sqrt2(\cos(\pi/4)+i\sin(\pi/4))\] \[(2\sqrt2)^6(\cos6(\pi/4)+i\sin6(\pi/4))\]

  29. amistre64
    • 4 years ago
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    tada!! :) i knew you could do it class is starting so ciao

  30. anonymous
    • 4 years ago
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    thanks! i kinda get it now.. but where does the 6 come from after the cos and sin

  31. EarthCitizen
    • 4 years ago
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    \[r=\sqrt{2^{2}+2^{2}}=\sqrt{4+4}=\sqrt{8}\] arg z =arctan(1)=45 \[(\sqrt{8}<45)^{6}\]

  32. anonymous
    • 4 years ago
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    oh wow alright thank you

  33. EarthCitizen
    • 4 years ago
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    \[r ^{n}<n \theta \] r=(8^1/2)^6=512, n=6 and theta=45*6=270 512<270\[r(\cos \theta+isin \theta) = 512(\cos(270)+isin(270))=-i512\]

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