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#4 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-02 01:22:57

Do you know how to combine impedances in parallel?

#5 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-02 01:17:47

No, post 1 was completely wrong. Sorry.

For the record, there is a Vieta's solution

#6 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-02 01:09:48

You can numerically do the problem now.

#7 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-02 00:39:14

try this

Table[(z - 2)^2015 == 2^2017 /. z -> (2^(2017/2015) Exp[I ((2 n \[Pi])/2015)] + 2), {n, 1,2015}]

#10 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-01 15:54:29

You can replace w with e^(i (2 n pi/ 2015)) where n is an integer from 1 to 2015 and i is the imaginary unit

#13 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-01 12:51:42

Okay. Do you see the rest of the solutions now?

#14 Re: Maths Teaching Resources » LearnMathsFree - Video series on geometry, topology, etc. » 2015-09-01 03:20:27

While I really think that this is an excellent effort, I would comment that you could have named your channel better

#15 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-01 03:00:08

When you take the nth root of something, you introduce n solutions.

For example, z^4 = 625 has roots z = 5w where w is the fourth root of unity. That is to say z = 5,-5,5i, -5i

Z-2)^2015 = 2^2017
Or, (Z-2) = (2^(2017/2015)) w
Or, Z = 2^(2017/2015) w + 2

The second line takes the 2015th root on both sides, wouldn't that introduce 2015 solutions?

#16 Re: Help Me ! » Complex Numbers in Electronics » 2015-09-01 00:18:40

That w is a complex root of unity. Do you know what that is?

#18 Re: Help Me ! » Complex Numbers in Electronics » 2015-08-31 12:00:31

I used M to do the sum. The answer is 10.

You are not doing the algebra right

(Z-2)^2015 = 2^2017
Or, (Z-2) = (2^(2017/2015)) w
Or, Z = 2^(2017/2015) w + 2

#20 Re: Help Me ! » Complex Numbers in Electronics » 2015-08-30 21:41:48

Why are you getting only one root, please show me your work

#22 Re: Help Me ! » Complex Numbers in Electronics » 2015-08-30 21:16:15

That can come later. The 4 is wrong, the 4 should be 2^(2017/2015).

What are the solutions to (Z-2)^2015 = 2^2017

#25 Re: Help Me ! » Complex Numbers in Electronics » 2015-08-30 11:58:18

The answer to the problem is not 6

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