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anonymous

  • one year ago

Write the complex number in the form a + bi. square root of six(cos 315° + i sin 315°)

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  1. anonymous
    • one year ago
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    I dont know how to solve this at all... My choices: square root of three over two minus square root of thee over two times i square root of six minus square root of sixi square root of three minus square root of threei square root of six over two minus square root of six over two times i

  2. Nnesha
    • one year ago
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    look at the unit circle cos 315 = what ? remember (x,y) sin =y-coordinate and cos = x-coordinate

  3. anonymous
    • one year ago
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    cos315 is square root(2)/2

  4. anonymous
    • one year ago
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    so \[\frac{ \sqrt{2} }{ 2 }\] is my x coordinate?

  5. Nnesha
    • one year ago
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    |dw:1439174169775:dw|

  6. anonymous
    • one year ago
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    Yes, I was just looking at that

  7. Nnesha
    • one year ago
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    \[\huge\rm \sqrt{6}(\cos 315° + i \sin 315°)\] so cos 315 =sqrt{2} over 2 replace cos 315 by that

  8. anonymous
    • one year ago
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    So i would replace isin315 with \[\frac{ -\sqrt{2} }{ 2 }\]

  9. Loser66
    • one year ago
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    @Nnesha Good job!! open my eye!! :)

  10. Nnesha
    • one year ago
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    \[\huge\rm \sqrt{6}(\color{reD}{cos 315°} + i \sin 315°)\] \[\sqrt{6}(\frac{ \sqrt{2} }{ 2 }+i (-\frac{ \sqrt{2} }{ 2 }))\] ust sin315 not the i

  11. Nnesha
    • one year ago
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    just**

  12. anonymous
    • one year ago
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    Alright, then do I distribute the \[\sqrt{6}\]?

  13. Nnesha
    • one year ago
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    thanks o^_^o @66

  14. Nnesha
    • one year ago
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    first distribute by i i times -sqrt{2} over 2

  15. Nnesha
    • one year ago
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    \[\sqrt{6}(\frac{ \sqrt{2} }{ 2} - \frac{ \sqrt{2} }{ 2}i)\]

  16. anonymous
    • one year ago
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    Okay, after distributing i, do I distribute the square root of 6?

  17. Nnesha
    • one year ago
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    well i wouldn't do that just take out the common factor

  18. Nnesha
    • one year ago
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    \[\sqrt{6}\color{ReD}{(\frac{ \sqrt{2} }{ 2} - \frac{ \sqrt{2} }{ 2}i)}\] what is common factor in the parentheses

  19. anonymous
    • one year ago
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    Hmm... then my end result is \[\sqrt{3}(-i+1)\]

  20. Nnesha
    • one year ago
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    u sure or just guessing ? ;P

  21. anonymous
    • one year ago
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    100% sure. I just calculated it. If I distribute the square root of 3, then I get what appears to be my third choice. Am I correct?

  22. Nnesha
    • one year ago
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    alright yes that's correct there are square roots that's why i don't like distributing by sqrt{6} you can take out the common factor which is sqrt{2}/2

  23. anonymous
    • one year ago
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    Oh, I see. @Nnesha

  24. Nnesha
    • one year ago
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    \[\sqrt{6} \times \frac{ \sqrt{2} }{ 2 }(1-i)\] \[\frac{ \sqrt{12} }{ 2 }(1-i)\]multply sqrt{6} times sqrt{2} now factor 12 \[\frac{ \sqrt{4 \times 3} }{ 2}(1-i)\] take the square root of 2 you will get the same answer

  25. Nnesha
    • one year ago
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    i found it easy but you can apply any method as lng s you get the same answer

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