Is the Textbook's Solution to the Complex Algebra Problem Incorrect?

[Plutonium88]

Homework Statement

http://postimg.org/image/tgx9b3n3t/

i drew a picture in paint of the initial question. Now me and my friend have both solved this question using different methods and have resulted in the same answer. The answer in the back of the book however is different from both of our answers. I'm wondering if some one can help me out and tell me if I'm missing something, or if the book is indeed wrong.

Homework Equations

(z^2 - z +1) / ((z^2 + 1/z^2)^2 + 2(z+1/z)^2 -3))^1/2

The Attempt at a Solution

Now what i did is i took the bottom piece and i subtracted and added 2 to make the first piece (z^2 + 1/z^2)^2 similar to the second piece

((z^2 + 1/z^2 +2 -2)^2 +2(z+1/z)^2 -3)^1/2

=(( (z+1/z)^2 - 2)^2 +2(z+1/z)^2 - 3)^1/2

when i simplify this i get

[ ((z+1/z)^2 - 1)^2 ] ^1/2

= z^2 - 2 + 1/z^2 -1

= (z^4 + z^2 + 1)/z^2

when i divide this piece by the initial numerator z^2 - z + 1

my answer yieldsz^2(z^2 -z +1)/ (z^4 + z^2 -1)The back of the books answer is

z^2/(z^2 + z +1)The only thing i could think of from my answer was that i could factor the bottom piece to cancel somethign with the top but the denominator is unfactorable from what i can see and what wolfram alpha can as well. sooo yea.. that's why i get stuck with my answer.

 

[DrClaude]

= (z^4 + z^2 + 1)/z^2

when i divide this piece by the initial numerator z^2 - z + 1

my answer yields


z^2(z^2 -z +1)/ (z^4 + z^2 -1)

Compare the first line and the last in the above quote.

 

[Plutonium88]

Compare the first line and the last in the above quote.

sorry i inverted that sentence.

It should say if i take the numerator divided by that peice.

ahh and it should say

z^2(z^2-z+1)/(z^4 + z^2 +1)

 

[haruspex]

sorry i inverted that sentence.

It should say if i take the numerator divided by that peice.

ahh and it should say

z^2(z^2-z+1)/(z^4 + z^2 +1)

... And can you see how to simplify that? Tip: to see whether polynomial P is a factor of Q, set P=0 and use that eqn to reduce Q progressively until it's of order less than that of P. P is a factor if and only if the result is 0.

 

[Plutonium88]

so I didn't use your hint because I don't believe I am understanding it correct, but what I saw was that if I take z^4 + z^2 +1 and I say z^4 +z^2 + 1 + z^2 - z^2 it will become
z^4 + 2z^2 + 1 - z^2
=(z^2 + 1)^2 - z^2

And this difference of squares will give you the factor. Alternatively my friend found you can multiply by (z+1)/(z+1) and this will also provide factors which will yield the answer. However I was curious if you could show me this factor method where 0=polynomial and you reduce the variable z in this case.

 

[SammyS]

Homework Statement

http://postimg.org/image/tgx9b3n3t/

i drew a picture in paint of the initial question. Now me and my friend have both solved this question using different methods and have resulted in the same answer. The answer in the back of the book however is different from both of our answers. I'm wondering if some one can help me out and tell me if I'm missing something, or if the book is indeed wrong.

Homework Equations

(z^2 - z +1) / ((z^2 + 1/z^2)^2 + 2(z+1/z)^2 -3))^1/2

The Attempt at a Solution

Now what i did is i took the bottom piece and i subtracted and added 2 to make the first piece (z^2 + 1/z^2)^2 similar to the second piece

((z^2 + 1/z^2 +2 -2)^2 +2(z+1/z)^2 -3)^1/2

=(( (z+1/z)^2 - 2)^2 +2(z+1/z)^2 - 3)^1/2

You appear to be saying that \displaystyle \ \left( z^2+\frac{1}{z^2}\right)^2=\left(\left(z+\frac{1}{z}\right)^2-2\right)^2\ ,\ which is correct.

Wow! That's easier to read in LaTeX !

when i simplify this i get

[ ((z+1/z)^2 - 1)^2 ] ^1/2

= z^2 - 2 + 1/z^2 -1

= (z^4 + z^2 + 1)/z^2

when i divide this piece by the initial numerator z^2 - z + 1

my answer yields


z^2(z^2 -z +1)/ (z^4 + z^2 -1)


The back of the books answer is

z^2/(z^2 + z +1)


The only thing i could think of from my answer was that i could factor the bottom piece to cancel something with the top but the denominator is unfactorable from what i can see and what wolfram alpha can as well. sooo yea.. that's why i get stuck with my answer.

Look at

\displaystyle \frac{z^2}{\displaystyle\left(\frac{z^4+z^2+1}{z^2-z+1}\right)}

using long division to simplify the denominator.

 

[haruspex]

I was curious if you could show me this factor method where 0=polynomial and you reduce the variable z in this case.

Q(z) = z⁴+z²+1; P(z)=z²-z+1
Setting P(z)=0 we have z² = z-1.
Substituting in Q:
Q(z) = z²(z²+1)+1 = (z-1)(z-1+1)+1 = z²-z+1 = P(z) = 0. Therefore P(z) divides Q(z).

 

[Plutonium88]

Q(z) = z⁴+z²+1; P(z)=z²-z+1
Setting P(z)=0 we have z² = z-1.
Substituting in Q:
Q(z) = z²(z²+1)+1 = (z-1)(z-1+1)+1 = z²-z+1 = P(z) = 0. Therefore P(z) divides Q(z).

Thank you sir.

 

[Plutonium88]

You appear to be saying that \displaystyle \ \left( z^2+\frac{1}{z^2}\right)^2=\left(\left(z+\frac{1}{z}\right)^2-2\right)^2\ ,\ which is correct.

Wow! That's easier to read in LaTeX !

Look at

\displaystyle \frac{z^2}{\displaystyle\left(\frac{z^4+z^2+1}{z^2-z+1}\right)}

using long division to simplify the denominator.

Okay so the method below to show that i can divide the function by the other function, and then using long division like you said. Thank you for your help