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

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

- freckles

and n is integer by the way

- anonymous

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

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## More answers

- freckles

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

- freckles

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

- freckles

do you know the unit circle?

- freckles

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

- freckles

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

- 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?

- freckles

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

- freckles

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

- freckles

since they are equal values

- anonymous

okay that makes sense, thank you very much

- freckles

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

- freckles

now we know the period is 2pi

- freckles

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

- freckles

this way will be more useful to us anyways

- freckles

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

- freckles

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

- anonymous

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

- freckles

no

- freckles

nothing should be divided by n

- freckles

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

- anonymous

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

- anonymous

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

- freckles

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

- anonymous

Yes, that makes sense

- freckles

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

- 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

- freckles

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

- 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

- 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

- freckles

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

- 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.

- 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

- 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

- anonymous

Yes each individual one

- freckles

omg that is so yucky

- 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

- freckles

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

- 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

- freckles

can you continue

- 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

- freckles

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

- anonymous

Yes it does thank you very much

- rock_mit182

is this complex calc ?

- anonymous

no this is actually a trigonometry class

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