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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
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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freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[z^8=i \\ \text{ hint } i=\cos(\frac{\pi}{2}+2n \pi)+ i \sin(\frac{\pi}{2}+2n \pi)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1and n is integer by the way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did you get that i=cos(π2+2nπ)+isin(π2+2nπ)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well it is easy to recall at pi/2 we have cos value is 0 and the sin value is 1

freckles
 one year ago
Best ResponseYou've already chosen the best response.1the +2n pi part comes from the number of rotations starting and ending at pi/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you know the unit circle?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1also I didn't get pi2 I got pi/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1since again cos(pi/2)=0 and sin(pi/2)=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why 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?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1yes and cos is 0 at pi/2 and sin is 1 at pi/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so you can replace 0 with cos(pi/2) and you replace 1 with sin(pi/2)

freckles
 one year ago
Best ResponseYou've already chosen the best response.1since they are equal values

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay that makes sense, thank you very much

freckles
 one year ago
Best ResponseYou've already chosen the best response.10+1i=cos(pi/2)+sin(pi/2)i since cos(pi/2)=0 and sin(pi/2)=1

freckles
 one year ago
Best ResponseYou've already chosen the best response.1now we know the period is 2pi

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so you could also write 0+1i=cos(pi/2+2npi)+sin(pi/2+2npi)i

freckles
 one year ago
Best ResponseYou've already chosen the best response.1this way will be more useful to us anyways

freckles
 one year ago
Best ResponseYou've already chosen the best response.1because we are going to need 8 answers n=0,1,2,3,4,5,6,7

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[z^8=e^{(\frac{\pi}{2}+2n \pi)i} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0shouldn't pi/2 be divided by n as well?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1nothing should be divided by n

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[z=e^{(\frac{\pi}{2(8)}+\frac{2n \pi}{8})i} \\ z=e^{(\frac{\pi}{16}+\frac{n \pi}{4})i}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[r^\frac{ 1 }{ n } = [\cos(\frac{ \theta+2\pi(k) }{ n })+isin(\frac{\theta+2\pi(k)}{n})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is the formula we need to use because we don't get a calculator on the exam

freckles
 one year ago
Best ResponseYou've already chosen the best response.1you do realize my n is acting as your k ? and that n you have there is your 8 for this question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, that makes sense

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I guess you aren't familiar with the e^ thingy notation \[e^{i \theta}=\cos(\theta)+i \sin(\theta) \\ \text{ way shorter \to write } e^{i \theta} \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Here 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1i went ahead and use the k and n as you used it above

freckles
 one year ago
Best ResponseYou've already chosen the best response.1here 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1anyways do you have any questions or have I rambled too much

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you have to write out all 8 solutions you know replace n with 0 then with 1 then with 2 ... then with 7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes each individual one

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[z_n=\cos(\frac{\pi}{16}+\frac{n \pi}{4})+i \sin(\frac{\pi}{16}+\frac{n \pi}{4}) \\ z_0=\cos(\frac{\pi}{16})+i \sin(\frac{\pi}{16}) \\ z_1=\cos(\frac{\pi}{16}+\frac{\pi}{4})+i \sin(\frac{\pi}{16}+\frac{\pi}{4}) \\ \text{ simplifying } z_1 \\ z_1=\cos(\frac{5\pi}{16})+i \sin(\frac{5\pi}{16})\] only 6 more to go... lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.1is that what you got when you plug in 1 though 5pi/16 for the theta part?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{\pi}{16}+\frac{n \pi}{4} \\ =\frac{\pi}{16}+\frac{4 n \pi}{16} \\ =\frac{\pi+4 n \pi}{16} \\ =\frac{\pi(1+4 n)}{16}\] 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.1np :) I'm glad you got it I hope this all makes more sense now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes it does thank you very much

rock_mit182
 one year ago
Best ResponseYou've already chosen the best response.0is this complex calc ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no this is actually a trigonometry class
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