Question

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

Answer 1

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

Answer 2

and n is integer by the way

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

do you know the unit circle?

Answer 7

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

Answer 8

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

Answer 9

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?

Answer 10

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

Answer 11

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

Answer 12

since they are equal values

Answer 13

okay that makes sense, thank you very much

Answer 14

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

Answer 15

now we know the period is 2pi

Answer 16

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

Answer 17

this way will be more useful to us anyways

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

no

Answer 22

nothing should be divided by n

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

Yes, that makes sense

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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)


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

Answer 33

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

Answer 34

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.

Answer 35

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

Answer 36

do you have to write out all 8 solutions
you know replace n with 0
then with 1
then with 2
...
then with 7

Answer 37

Yes each individual one

Answer 38

omg that is so yucky

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

can you continue

Answer 43

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

Answer 44

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

Answer 45

Yes it does thank you very much

Answer 46

is this complex calc ?

Answer 47

no this is actually a trigonometry class