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.

$\displaystyle \Large (z-1)^6=(z+1)^6$

I'm not supposed to mulitply anything out.

$\displaystyle \Large 1=\ (\frac{1-z}{1+z})^6$

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.

$\displaystyle -1^{even exponent}=1$ ... always. 6 is obviously an even exponent.

The reason for this is that:

$\displaystyle 1^6=-1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \lor 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1$

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: $\displaystyle \sqrt{x^2} = \pm x $

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.

$\displaystyle -1^{even exponent}=1$ ... always. 6 is obviously an even exponent.

The reason for this is that:

$\displaystyle 1^6=-1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \lor 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1$

The general rule: $\displaystyle \sqrt{x^2} = \pm x $

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

First The general rule: $\displaystyle \sqrt{x^2} \ne \pm x $ !!!

This is true: The general rule: $\displaystyle \sqrt{x^2} = |x| $

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 $\displaystyle \sqrt{4}= \pm 2$, as both $\displaystyle 2^2$ and $\displaystyle (-2)^2$ would yield $\displaystyle 4$. 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 $\displaystyle \sqrt{4}= \pm 2$, as both $\displaystyle 2^2$ and $\displaystyle (-2)^2$ would yield $\displaystyle 4$. 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.

$\displaystyle \sqrt{4}=2~\&~\sqrt{4}\ne -2$.

If the question were what are the square roots of $\displaystyle 16$ then the answer would be $\displaystyle \pm 4$.

However the symbol $\displaystyle \sqrt{16}=4$.

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 $\displaystyle -x^n=y$, 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 $\displaystyle \left(\frac{1- z}{1+ z}\right)^6= 1$, note that the 6 6th roots of 1 are $\displaystyle 1$, $\displaystyle cos(\pi/3)+ i sin(\pi/3)= 1/2+ i\sqrt{3}/2$, $\displaystyle cos(2\pi/3)+ i sin(2\pi/3)= -1/2+ i\sqrt{3}/2$, $\displaystyle -1$, $\displaystyle -1/2- i\sqrt{3}/2$, and $\displaystyle -1/2+ i\sqrt{3}/2$.

Set $\displaystyle \frac{1- z}{1+ z}$ equal to each of those and solve the resulting linear equation.