1. Apr 22, 2012

### ronho1234

let A={z|z^6=√3 + i} B=(z|Im(z)>0} and C={z|Re(z)>0} find A∩B∩C
the part previous to this qn asks me to find the roots of z^6 and ive already down that. but i have no idea how to proceed with this, so do i draw my unit ciorcle with the hexagon and then follow to see what regions satisfies with the other 2? please help

2. Apr 22, 2012

### Mentallic

The sets B and C are just restricting you to pick all the roots from A such that it lies in the first quadrant. Can you figure out which ones these are?

Oh and it roots don't lie on the unit circle, because the unit circle has has modulus of 1 and these roots have a different modulus.

3. Apr 22, 2012

### Curious3141

So what values of z did you find in the first part? Just identify the ones with positive real *and* positive imaginary parts in those and put them in curly brackets {} separated by commas.

4. Apr 22, 2012

### ronho1234

umm ok so for solving z^6=root3 + i
z=2^1/6 cis(pi/36 +k2pi/6)
the roots are for k=0,1,2,3,4,5 2^1/6cis[pi/36 or 13pi/36 or 25pi/36 or 37pi/36 or 49pi/36 or 62pi/36]

so i find out that its the first two value 2^1/6cis [pi/36 and 13pi/36] that are common with the other 2 statements

so for the answer do i just put these 2 values in the swirly brackets as the answer, umm im not sure....

5. Apr 22, 2012

### Curious3141

Yes, but put them in a+bi form.

6. Apr 22, 2012

### ronho1234

ummm because i get wacky values for the arguments its really hard or i just can't seem to switch them back to a+bi form i get decimals i think....

7. Apr 22, 2012

### Curious3141

Just to clarify, I meant leave them as trig ratios, but express them as r(cosθ +i sinθ) rather than using the shorthand "cis".

8. Apr 22, 2012

### ronho1234

oh sorry!... so just
A∩B∩C = {2^1/6(cos(pi/36)+isin(pi/36),2^1/6(cos(13pi/36)+isin(13pi/36)}

9. Apr 22, 2012

### Curious3141

Yeah, I'm sure it'll come out better when you write it down, but that's the idea.

10. Apr 22, 2012

### ronho1234

ok THANKYOU curious3141 and mentallic :)

11. Apr 22, 2012

### Mentallic

Just as a heads up, since $r\cdot cis(\theta)\equiv r(\cos(\theta)+i\sin(\theta))$ then it should be ok to leave the answer in cis form, because that is it's definition simply because it's short and neat.

I'd only ever separate it when it makes sense to separate it such as when you want to show why if $z=cis\theta$ then $\overline{z}=cis(-\theta)$ for example.

12. Apr 22, 2012

### ronho1234

ok thanks for the extra note, i'll keep that in mind