## 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

$z^8=i \\ \text{ hint } i=\cos(\frac{\pi}{2}+2n \pi)+ i \sin(\frac{\pi}{2}+2n \pi)$

2. freckles

and n is integer by the way

3. anonymous

How did you get that i=cos(π2+2nπ)+isin(π2+2nπ)

4. freckles

well it is easy to recall at pi/2 we have cos value is 0 and the sin value is 1

5. freckles

the +2n pi part comes from the number of rotations starting and ending at pi/2

6. freckles

do you know the unit circle?

7. freckles

also I didn't get pi2 I got pi/2

8. freckles

since again cos(pi/2)=0 and sin(pi/2)=1

9. anonymous

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

yes and cos is 0 at pi/2 and sin is 1 at pi/2

11. freckles

so you can replace 0 with cos(pi/2) and you replace 1 with sin(pi/2)

12. freckles

since they are equal values

13. anonymous

okay that makes sense, thank you very much

14. freckles

0+1i=cos(pi/2)+sin(pi/2)i since cos(pi/2)=0 and sin(pi/2)=1

15. freckles

now we know the period is 2pi

16. freckles

so you could also write 0+1i=cos(pi/2+2npi)+sin(pi/2+2npi)i

17. freckles

this way will be more useful to us anyways

18. freckles

because we are going to need 8 answers n=0,1,2,3,4,5,6,7

19. freckles

$z^8=e^{(\frac{\pi}{2}+2n \pi)i}$

20. anonymous

shouldn't pi/2 be divided by n as well?

21. freckles

no

22. freckles

nothing should be divided by n

23. freckles

$z=e^{(\frac{\pi}{2(8)}+\frac{2n \pi}{8})i} \\ z=e^{(\frac{\pi}{16}+\frac{n \pi}{4})i}$

24. anonymous

$r^\frac{ 1 }{ n } = [\cos(\frac{ \theta+2\pi(k) }{ n })+isin(\frac{\theta+2\pi(k)}{n})$

25. anonymous

This is the formula we need to use because we don't get a calculator on the exam

26. freckles

you do realize my n is acting as your k ? and that n you have there is your 8 for this question

27. anonymous

Yes, that makes sense

28. freckles

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

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

i went ahead and use the k and n as you used it above

31. freckles

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

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

anyways do you have any questions or have I rambled too much

34. anonymous

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

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

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

Yes each individual one

38. freckles

omg that is so yucky

39. freckles

$z_n=\cos(\frac{\pi}{16}+\frac{n \pi}{4})+i \sin(\frac{\pi}{16}+\frac{n \pi}{4}) \\ z_0=\cos(\frac{\pi}{16})+i \sin(\frac{\pi}{16}) \\ z_1=\cos(\frac{\pi}{16}+\frac{\pi}{4})+i \sin(\frac{\pi}{16}+\frac{\pi}{4}) \\ \text{ simplifying } z_1 \\ z_1=\cos(\frac{5\pi}{16})+i \sin(\frac{5\pi}{16})$ only 6 more to go... lol

40. freckles

is that what you got when you plug in 1 though 5pi/16 for the theta part?

41. freckles

$\frac{\pi}{16}+\frac{n \pi}{4} \\ =\frac{\pi}{16}+\frac{4 n \pi}{16} \\ =\frac{\pi+4 n \pi}{16} \\ =\frac{\pi(1+4 n)}{16}$ 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

can you continue

43. anonymous

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

np :) I'm glad you got it I hope this all makes more sense now

45. anonymous

Yes it does thank you very much

46. rock_mit182

is this complex calc ?

47. anonymous

no this is actually a trigonometry class