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anonymous

  • one year ago

Solve the equation. (List your answers counterclockwise about the origin starting at the positive real axis. Express θ in radians.) z^8 − i = 0

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  1. freckles
    • one year ago
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    \[z^8=i \\ \text{ hint } i=\cos(\frac{\pi}{2}+2n \pi)+ i \sin(\frac{\pi}{2}+2n \pi)\]

  2. freckles
    • one year ago
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    and n is integer by the way

  3. anonymous
    • one year ago
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    How did you get that i=cos(π2+2nπ)+isin(π2+2nπ)

  4. freckles
    • one year ago
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    well it is easy to recall at pi/2 we have cos value is 0 and the sin value is 1

  5. freckles
    • one year ago
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    the +2n pi part comes from the number of rotations starting and ending at pi/2

  6. freckles
    • one year ago
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    do you know the unit circle?

  7. freckles
    • one year ago
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    also I didn't get pi2 I got pi/2

  8. freckles
    • one year ago
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    since again cos(pi/2)=0 and sin(pi/2)=1

  9. anonymous
    • one year ago
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    why do you need them to equal 0 and 1? that's what i'm not understanding. \[z^8=i \] in a+bi form is 0+1i right?

  10. freckles
    • one year ago
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    yes and cos is 0 at pi/2 and sin is 1 at pi/2

  11. freckles
    • one year ago
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    so you can replace 0 with cos(pi/2) and you replace 1 with sin(pi/2)

  12. freckles
    • one year ago
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    since they are equal values

  13. anonymous
    • one year ago
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    okay that makes sense, thank you very much

  14. freckles
    • one year ago
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    0+1i=cos(pi/2)+sin(pi/2)i since cos(pi/2)=0 and sin(pi/2)=1

  15. freckles
    • one year ago
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    now we know the period is 2pi

  16. freckles
    • one year ago
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    so you could also write 0+1i=cos(pi/2+2npi)+sin(pi/2+2npi)i

  17. freckles
    • one year ago
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    this way will be more useful to us anyways

  18. freckles
    • one year ago
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    because we are going to need 8 answers n=0,1,2,3,4,5,6,7

  19. freckles
    • one year ago
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    \[z^8=e^{(\frac{\pi}{2}+2n \pi)i} \]

  20. anonymous
    • one year ago
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    shouldn't pi/2 be divided by n as well?

  21. freckles
    • one year ago
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    no

  22. freckles
    • one year ago
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    nothing should be divided by n

  23. freckles
    • one year ago
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    \[z=e^{(\frac{\pi}{2(8)}+\frac{2n \pi}{8})i} \\ z=e^{(\frac{\pi}{16}+\frac{n \pi}{4})i}\]

  24. anonymous
    • one year ago
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    \[r^\frac{ 1 }{ n } = [\cos(\frac{ \theta+2\pi(k) }{ n })+isin(\frac{\theta+2\pi(k)}{n})\]

  25. anonymous
    • one year ago
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    This is the formula we need to use because we don't get a calculator on the exam

  26. freckles
    • one year ago
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    you do realize my n is acting as your k ? and that n you have there is your 8 for this question

  27. anonymous
    • one year ago
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    Yes, that makes sense

  28. freckles
    • one year ago
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    I guess you aren't familiar with the e^ thingy notation \[e^{i \theta}=\cos(\theta)+i \sin(\theta) \\ \text{ way shorter \to write } e^{i \theta} \]

  29. freckles
    • one year ago
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    Here is something more general: Say we want to solve: \[z^n= a+bi \] \[ \text{ First write } a+bi \text{ in polar form } \] \[r=\sqrt{a^2+b^2} \\ \text{ now if } a,b>0 \text{ then } \theta=\arctan(\frac{b}{a}) \\ \text{ if } a<0,b>0 \text{ then } \theta=\arctan(\frac{b}{a})+\pi \\ \text{ if } a,b<0 \text{ then } \theta=\arctan(\frac{b}{a})+\pi \\ \text{ if } a>0,b<0 \text{ then } \theta=\arctan(\frac{b}{a}) \\ \text{ so \let's go for one these \cases } \\ \text{ \let's assume looking at } z^n=a+bi \\ \text{ we have the case } a,b>0 \\ \text{ so } z^n=r(\cos(\arctan(\frac{b}{a})+2k \pi)+i \sin(\frac{b}{a})+2 k \pi)) \\ \text{ now } z=r^\frac{1}{n}(\cos(\frac{1}{n} \arctan(\frac{b}{a})+\frac{2 k \pi}{n})+i \sin( \frac{1}{n} \arctan(\frac{b}{a})+\frac{2 k \pi}{n} ))\] now I know that probably looks hideous to you it would probably look even more hideous if I did that 2nd or 3rd case

  30. freckles
    • one year ago
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    i went ahead and use the k and n as you used it above

  31. freckles
    • one year ago
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    here is another example: \[z^3=-1+i \\ r=\sqrt{(-1)^2+(1)^2}=\sqrt{1+1}=\sqrt{2} \\ \text{ since } a=-1 \text{ and } b=1 \\ \text {we have } \theta=\arctan(\frac{1}{-1})+\pi=\arctan(-1)+\pi=\frac{-\pi}{4}+\pi =\frac{3\pi}{4} \\ \text{ so } z^3=\sqrt{2}(\cos(\frac{3\pi}{4}+2 k \pi)+i \sin(\frac{3\pi}{4} +2k \pi)) \\ \text{ now we find } z \\ z=(\sqrt{2})^\frac{1}{3}(\cos(\frac{3\pi}{4(3)}+\frac{2k \pi}{3})+i \sin(\frac{3\pi}{4(3)}+\frac{2 k \pi}{3}))\] where we will take k=0 and k=1 and then finally k=2 to find out 3 solutions

  32. freckles
    • one year ago
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    I also remember that in the 1st and 4th quadrant is where arctan( ) will output appropriate values because recall the range of arctan( ) is -pi/2 to pi/2 which includes only the 1st and 4th quadrant but to get to the other quadrants the 2nd and 3rd I just remember to add pi (you could subtract pi or some odd integer*pi) |dw:1443894443450:dw| it helps to visualize where a+bi actually is to determine whether to just take the output arctan(b/a) or to take the output arctan(b/a)+pi for theta

  33. freckles
    • one year ago
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    anyways do you have any questions or have I rambled too much

  34. anonymous
    • one year ago
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    I'm using a webassign virtual homework and it keeps marking the answer incorrect although I know I have this correct at this point. I'm not sure if it's the program that's messing up or I'm still doing something wrong.

  35. freckles
    • one year ago
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    so based on your description in the instructions we don't need to use a calculator since it asked to write theta in radians you do have this: \[z=e^{(\frac{\pi}{2(8)}+\frac{2n \pi}{8})i} \\ z=e^{(\frac{\pi}{16}+\frac{n \pi}{4})i}\] or I mean in your class I think you are just using that one long form so this: \[z=\cos(\frac{\pi}{16}+\frac{ n \pi}{4})+i \sin(\frac{\pi}{16}+\frac{n \pi}{4})\] and maybe since your class is using k as the integer instead of n replace my n with k

  36. freckles
    • one year ago
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    do you have to write out all 8 solutions you know replace n with 0 then with 1 then with 2 ... then with 7

  37. anonymous
    • one year ago
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    Yes each individual one

  38. freckles
    • one year ago
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    omg that is so yucky

  39. freckles
    • one year ago
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    \[z_n=\cos(\frac{\pi}{16}+\frac{n \pi}{4})+i \sin(\frac{\pi}{16}+\frac{n \pi}{4}) \\ z_0=\cos(\frac{\pi}{16})+i \sin(\frac{\pi}{16}) \\ z_1=\cos(\frac{\pi}{16}+\frac{\pi}{4})+i \sin(\frac{\pi}{16}+\frac{\pi}{4}) \\ \text{ simplifying } z_1 \\ z_1=\cos(\frac{5\pi}{16})+i \sin(\frac{5\pi}{16})\] only 6 more to go... lol

  40. freckles
    • one year ago
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    is that what you got when you plug in 1 though 5pi/16 for the theta part?

  41. freckles
    • one year ago
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    \[\frac{\pi}{16}+\frac{n \pi}{4} \\ =\frac{\pi}{16}+\frac{4 n \pi}{16} \\ =\frac{\pi+4 n \pi}{16} \\ =\frac{\pi(1+4 n)}{16}\] there might make it easier combining the fractions before pluggin in all of that let's just play with theta we know what form it needs to go in so for n=0 the theta is pi/16 for n=1 the theta is 5pi/16 for n=2 the theta is 9pi/16 for n=3 the theta is 13pi/16

  42. freckles
    • one year ago
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    can you continue

  43. anonymous
    • one year ago
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    Yes I finally got this, so I was getting the right answer now but I had a syntax error in the program that was making it sensitive. Thank you so much for your help

  44. freckles
    • one year ago
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    np :) I'm glad you got it I hope this all makes more sense now

  45. anonymous
    • one year ago
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    Yes it does thank you very much

  46. rock_mit182
    • one year ago
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    is this complex calc ?

  47. anonymous
    • one year ago
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    no this is actually a trigonometry class

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