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EarthCitizen

  • 4 years ago

((3+j3)^3*(1-j)^4)/((1+j(3)^1/2)^9)

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  1. EarthCitizen
    • 4 years ago
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    \[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}\]

  2. EarthCitizen
    • 4 years ago
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    yo

  3. asnaseer
    • 4 years ago
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    Just digesting that equation :)

  4. asnaseer
    • 4 years ago
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    ok - it would be best to try and break it down a bit to simplify it first...

  5. asnaseer
    • 4 years ago
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    \[(3+3j)^3=(3(1+j))^3=3^3(1+j)^3=27(1+j)^3\]

  6. asnaseer
    • 4 years ago
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    next simplification...

  7. asnaseer
    • 4 years ago
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    \[(3+3j)^3*(1-j)^4=27(1+j)^3*(1-j)^4=27*((1+j)(1-j))^3(1-j)\]\[=27*(1-j^2)^3(1-j)=27*(1+1)^3(1-j)=27*2^3(1-j)\]\[=27*8(1-j)=216(1-j)\]

  8. asnaseer
    • 4 years ago
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    make sense so far?

  9. EarthCitizen
    • 4 years ago
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    yep

  10. asnaseer
    • 4 years ago
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    ok, so now we are left with:\[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1-j)}{(1+j\sqrt{3})^9}\]

  11. asnaseer
    • 4 years ago
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    next lets try and simplify the denominator...

  12. EarthCitizen
    • 4 years ago
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    alryt

  13. asnaseer
    • 4 years ago
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    \[(1+j\sqrt{3})^2=1+j2\sqrt{3}+3j^2=1+j2\sqrt{3}-3=-2+j2\sqrt{3}=-2(1-j\sqrt{3})\]therefore:\[(1+j\sqrt{3})^3=(1+j\sqrt{3})^2(1+j\sqrt{3})=-2(1-j\sqrt{3})(1+j\sqrt{3})\]\[=-2(1-j^23)=-2(1+3)=-2*4=-8\]

  14. asnaseer
    • 4 years ago
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    therefore:\[(1+j\sqrt{3})^9=((1+j\sqrt{3})^3)^3=(-8)^3=-512\]

  15. asnaseer
    • 4 years ago
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    can you complete the rest now?

  16. EarthCitizen
    • 4 years ago
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    uhmm.., cpould you plz divide the top by the bottom for a final solution ?

  17. asnaseer
    • 4 years ago
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    \[((3+j3)^{3} *(1-j)^{4})/(1+j \sqrt{3})^{9}=\frac{216(1-j)}{(1+j\sqrt{3})^9}=\frac{216(1-j)}{-512}=-\frac{27(1-j)}{64}\]

  18. EarthCitizen
    • 4 years ago
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    kwl, your awesome!

  19. asnaseer
    • 4 years ago
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    yw - the main point however is that you understand the process - I hope you did :)

  20. asnaseer
    • 4 years ago
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    the "key" I guess is spotting how to get the equation into a form like: \((a+jb)(a-jb)\) so that it can be simplified to \((a+jb)(a-jb)=a^2+b^2\)

  21. asnaseer
    • 4 years ago
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    that will come with practice.

  22. EarthCitizen
    • 4 years ago
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    yh, hang on. so you used difference of two squares only ?

  23. asnaseer
    • 4 years ago
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    sum of two squares

  24. asnaseer
    • 4 years ago
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    \[a^2-b^2=(a+b)(a-b)\]\[a^2+b^2=(a+jb)(a-jb)\]

  25. EarthCitizen
    • 4 years ago
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    yh, so why didn't you use de Moivre's theorem ?

  26. asnaseer
    • 4 years ago
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    you could, but then you would need to find the correct angle for:\[e^{j\theta}=\cos(\theta)+j\sin(\theta)\]and I /feel/ the approach I used instead is often much simpler.

  27. EarthCitizen
    • 4 years ago
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    yh. a lot simpler coz it started getting cloudy , thanx tho

  28. asnaseer
    • 4 years ago
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    np - you are more than welcome. I'm glad I was able to help.

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