Solving z^6 and Finding A∩B∩C: A Comprehensive Guide for Beginners

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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 I've 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
 
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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.
 
ronho1234 said:
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 I've 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

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.
 
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 I am not sure...
 
ronho1234 said:
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 I am not sure...

Yes, but put them in a+bi form.
 
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...
 
ronho1234 said:
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...

Just to clarify, I meant leave them as trig ratios, but express them as r(cosθ +i sinθ) rather than using the shorthand "cis".
 
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)}
would be my final answer??
 
ronho1234 said:
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)}
would be my final answer??

Yeah, I'm sure it'll come out better when you write it down, but that's the idea.
 
ok THANKYOU curious3141 and mentallic :)
 
Just as a heads up, since [itex]r\cdot cis(\theta)\equiv r(\cos(\theta)+i\sin(\theta))[/itex] 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 [itex]z=cis\theta[/itex] then [itex]\overline{z}=cis(-\theta)[/itex] for example.
 
ok thanks for the extra note, i'll keep that in mind