SuprDooprPoopr
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I'm not sure how to begin doing this question
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williamnguyen
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write 1 in cis form (1 =cis(0) )
use de-moivre theorem to find your solutions in cis form then u can write it in a+bi form. Good luck!
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SuprDooprPoopr
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(Original post by williamnguyen)
write 1 in cis form (1 =cis(0) )
use de-moivre theorem to find your solutions in cis form then u can write it in a+bi form. Good luck!
I'm sorry, I'm a bit behind on complex numbers... how can I write 1 in cis form? And how do I resolve z^6?
I can use De-Moivre's theorem.
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williamnguyen
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(Original post by SuprDooprPoopr)
I'm sorry, I'm a bit behind on complex numbers... how can I write 1 in cis form? And how do I resolve z^6?
I can use De-Moivre's theorem.
Okay, so if u plot 1 on the real and imaginary axis, it is located 1 unit horizontally right. So the magnitude/length is 1 and the angle is 0 radian
--> 1 = cos(0) + i sin(0) = cis(0) = z^6 (in fact the angle can be 2pi, 4pi, and so on every time u go around the circle. The angle is in the form of 2kpi where k is an integer)
If you raise both sides to the power of 1/6,
z = (cis(0+2kpi) )^1/6
Using de-movire theorem,
z = cis(k*pi/3). (you just multiply the angle inside by 1/6)
You will get more than more solutions. Use k = 0,1,2, ... 5
--> 6 solutions in total. VAMOS!
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SuprDooprPoopr
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(Original post by williamnguyen)
Okay, so if u plot 1 on the real and imaginary axis, it is located 1 unit horizontally right. So the magnitude/length is 1 and the angle is 0 radian
--> 1 = cos(0) + i sin(0) = cis(0) = z^6 (in fact the angle can be 2pi, 4pi, and so on every time u go around the circle. The angle is in the form of 2kpi where k is an integer)
If you raise both sides to the power of 1/6,
z = (cis(0+2kpi) )^1/6
Using de-movire theorem,
z = cis(k*pi/3). (you just multiply the angle inside by 1/6)
You will get more than more solutions. Use k = 0,1,2, ... 5
--> 6 solutions in total. VAMOS!
thank you, i can now apply this technique to similar problems
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RDKGames
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(Original post by SuprDooprPoopr)
I'm sorry, I'm a bit behind on complex numbers... how can I write 1 in cis form? And how do I resolve z^6?
I can use De-Moivre's theorem.
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SuprDooprPoopr
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(Original post by williamnguyen)
Okay, so if u plot 1 on the real and imaginary axis, it is located 1 unit horizontally right. So the magnitude/length is 1 and the angle is 0 radian
--> 1 = cos(0) + i sin(0) = cis(0) = z^6 (in fact the angle can be 2pi, 4pi, and so on every time u go around the circle. The angle is in the form of 2kpi where k is an integer)
If you raise both sides to the power of 1/6,
z = (cis(0+2kpi) )^1/6
Using de-movire theorem,
z = cis(k*pi/3). (you just multiply the angle inside by 1/6)
You will get more than more solutions. Use k = 0,1,2, ... 5
--> 6 solutions in total. VAMOS!
One more question, why use k = 0,1,2, ... 5? The question asks for all z : C. Thank you for the help so far

edit: I understand, I have all the possible solutions that it could be.. thank you!
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SuprDooprPoopr
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(Original post by RDKGames)
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Thank you, I understand how to do the question now, just figuring out the last bit in my last message
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RDKGames
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(Original post by SuprDooprPoopr)
One more question, why use k = 0,1,2, ... 5? The question asks for all z : C. Thank you for the help so far

edit: I understand, I have all the possible solutions that it could be.. thank you!
Because when k is anything else, then you start finding solutions you already got. Try k=6.
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