# Complex Number Question

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#2

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!

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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(Original post by

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!

**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 can use De-Moivre's theorem.

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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.

**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.

--> 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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(Original post by

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!

**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!

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#6

**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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**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!

edit: I understand, I have all the possible solutions that it could be.. thank you!

Last edited by SuprDooprPoopr; 1 month ago

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**RDKGames**)
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(Original post by

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!

**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!

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