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

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January 20th 2013, 07:37 AM
director

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

I'm not supposed to mulitply anything out.
$\Large 1=\ (\frac{1-z}{1+z})^6$

I think this is right. What can I do next?
Can I simplify this further? Subsitute z for the polar form?
Thanks
January 20th 2013, 08:58 AM
JacobOCD

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

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

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

The reason for this is that:

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

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

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

(Wink)
January 20th 2013, 09:09 AM
JacobOCD

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)
January 20th 2013, 09:24 AM
Plato

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.
$-1^{even exponent}=1$ ... always. 6 is obviously an even exponent.
The reason for this is that:
$1^6=-1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \cdot -1 \lor 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1$

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

There are several mistakes in that reply.

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

This is true: The general rule: $\sqrt{x^2} = |x|$
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.
January 20th 2013, 09:30 AM
JacobOCD

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 $\sqrt{4}= \pm 2$ , as both $2^2$ and $(-2)^2$ would yield $4$ . I know that this is somewhat irrelevant, but I would like to know what is "wrong".
January 20th 2013, 09:46 AM
Plato

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

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

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

However the symbol $\sqrt{16}=4$ .
January 20th 2013, 09:57 AM
JacobOCD

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 $-x^n=y$ , y being positive.
January 20th 2013, 12:02 PM
HallsofIvy

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 roots of 16" (note the plural) are -4 and 4.

director, if you are not allowed to multiply out, then, starting from $\left(\frac{1- z}{1+ z}\right)^6= 1$ , note that the 6 6th roots of 1 are $1$ , $cos(\pi/3)+ i sin(\pi/3)= 1/2+ i\sqrt{3}/2$ , $cos(2\pi/3)+ i sin(2\pi/3)= -1/2+ i\sqrt{3}/2$ , $-1$ , $-1/2- i\sqrt{3}/2$ , and $-1/2+ i\sqrt{3}/2$ .

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