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Meaning?

Do you know how to combine impedances in parallel?

No, post 1 was completely wrong. Sorry.

For the record, there is a Vieta's solution

You can numerically do the problem now.

try this

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

There is a mistake with that code

There is a Vieta's solution

Yes, see post 35.

Meaning?

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

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?

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

Try guessing

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

Mechanically?

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

there are obviously 2015 roots?

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

The 4 is wrong

The answer to the problem is not 6

What! Why?

Can you formulate the question yourself?