find zeros of (1-z)^6 = (1+z)^6

Hi all,

I don't know if this is the right forum.

I have the following equation and I need to find all the solutions in C.

I'm not supposed to mulitply anything out.

I think this is right. What can I do next?

Can I simplify this further? Subsitute z for the polar form?

Thanks

Re: find zeros of (1-z)^6 = (1+z)^6

This is a trick which is supposed to make you think about exponents.

... always. 6 is obviously an even exponent.

The reason for this is that:

The opposite is true with an uneven value in the exponent: Then you don't have a minus for every other minus.

The general rule:

So basically you can probably think out the solution from there, I think.

(Wink)

Re: find zeros of (1-z)^6 = (1+z)^6

LOL, sorry, I misread your post. I don't think my answer is relevant. (Rofl)

Re: find zeros of (1-z)^6 = (1+z)^6

Quote:

Originally Posted by

**JacobOCD** This is a trick which is supposed to make you think about exponents.

... always. 6 is obviously an even exponent.

The reason for this is that:

The general rule:

**There are several mistakes in that reply.**

First The general rule: !!!

This is true: The general rule:

I don't think that it is a trick question at all.

As to the OP, I don't understand why "I'm not supposed to multiply anything out."

As you can see here there are five solutions.

Re: find zeros of (1-z)^6 = (1+z)^6

I forgot. What does your notation imply? The same as what I meant? The thing that I tried to imply was that the possible solution to for instance , as both and would yield . I know that this is somewhat irrelevant, but I would like to know what is "wrong".

Re: find zeros of (1-z)^6 = (1+z)^6

Quote:

Originally Posted by

**JacobOCD** I forgot. What does your notation imply? The same as what I meant? The thing that I tried to imply was that the possible solution to for instance

, as both

and

would yield

. I know that this is somewhat irrelevant, but I would like to know what is "wrong".

If you are going to try and help, then you need to know the basics.

.

If the question were what are the square roots of then the answer would be .

However the symbol .

Re: find zeros of (1-z)^6 = (1+z)^6

"Square roots of 16" isn't the same as 16 under a square root symbol? But yeah, I do get your point, I would actually define the square roots of 4 and 16 as 2 and 4, but there is always the at least theoretical posibility of the value coming to be on the basis of a *n*, even quantity of negative numbers multiplied together, giving , y being positive.

Re: find zeros of (1-z)^6 = (1+z)^6

No, it's not. "16 under a square root symbol" is 4, by definition of "square root". The "square root**s** of 16" (note the plural) are -4 and 4.

director, if you are not allowed to multiply out, then, starting from , note that the 6 6th roots of 1 are , , , , , and .

Set equal to each of those and solve the resulting linear equation.