1. ## [SOLVED] Complex numbers

If $z=e^{x}$, then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=$

$x=\frac{2\pi\mathbf{i}}{5}$

I am thinking this might be able to be represented in a series but the coefficients are making that an issue. Maybe there is another way to tackle this?

I put this in this form because it wasn't necessarily to calculus based but it isn't exactly pre-calculus.

2. Originally Posted by dwsmith
If $z=e^{x}$, then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=$

$x=\frac{2\pi\mathbf{i}}{5}$

I am thinking this might be able to be represented in a series but the coefficients are making that an issue. Maybe there is another way to tackle this?

I put this in this form because it wasn't necessarily to calculus based but it isn't exactly pre-calculus.

$1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=$ $\sum^9_{n=0}z^n+z^4\left(4+3z+3z^2+3z^3+3z^4+4z^5\ right)$ $=3z^4\sum^5_{n=0}z^n+z^4+z^5=$

$=3z^4\,\frac{z^6-1}{z-1}+z^4+z^5=$ $3e^{-2\pi i/5}\,\frac{e^{2\pi i/5}-1}{e^{2\pi i/5-1}}+e^{-2\pi i/5}+1$ $=e^{-2\pi i/5}+1$

Things to take into account: $\sum^k_{n=0}a^k=\frac{a^{k+1}-1}{a-1}\,,\,\,e^{k\pi i}=\left\{\begin{array}{ll}1&,\,\,if\,\,k\in\mathb b{Z}\,\,is\,\,even\\\!\!\!\!-1&,\,\,if\,\,k\in\mathbb{Z}\,\,is\,\,odd\end{array }\right.$

Check carefully the above, some mistakes may have snuck in.

Tonio

3. Originally Posted by tonio
$1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=$ $\sum^9_{n=0}z^n+z^4\left(4+3z+3z^2+3z^3+3z^4+4z^5\ right)$ $=3z^4\sum^5_{n=0}z^n+z^4+z^5=$

$=3z^4\,\frac{z^6-1}{z-1}+z^4+z^5=$ $3e^{-2\pi i/5}\,\frac{e^{2\pi i/5}-1}{e^{2\pi i/5-1}}+e^{-2\pi i/5}+1$ $=e^{-2\pi i/5}+1$

Things to take into account: $\sum^k_{n=0}a^k=\frac{a^{k+1}-1}{a-1}\,,\,\,e^{k\pi i}=\left\{\begin{array}{ll}1&,\,\,if\,\,k\in\mathb b{Z}\,\,is\,\,even\\\!\!\!\!-1&,\,\,if\,\,k\in\mathbb{Z}\,\,is\,\,odd\end{array }\right.$

Check carefully the above, some mistakes may have snuck in.

Tonio
I don't understand how you came up with that sum but I am pretty sure it is incorrect since your answer differs from the actual answer, $-5e^{\frac{3\pi\mathbf{i}}{5}}$.

4. Originally Posted by dwsmith
If $z=e^{x}$, then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=$

$x=\frac{2\pi\mathbf{i}}{5}$

I am thinking this might be able to be represented in a series but the coefficients are making that an issue. Maybe there is another way to tackle this?

I put this in this form because it wasn't necessarily to calculus based but it isn't exactly pre-calculus.
$z^5= e^{2\pi i}= 1$ so this is

$1+ z+ z^2+z^3+ 5z^4+ 4+ 4z+ 4z^2+ 4z^3+ 5z^4= 5+ 5z^2+ 5z^3+ 5z^4= 5(1+ z+ z^2+ z^3+ z^4)$.

And, since $z^5= 1$, $z^5- 1= (z- 1)(z^4+ z^3+ z^2+ z+ 1)= 0$ and since $z- 1\ne 0$ for $z= e^{i\pi/5}$.....

No, the answer is NOT $
-5e^{\frac{3\pi\mathbf{i}}{5}}
$

5. Originally Posted by HallsofIvy
$z^5= e^{2\pi i}= 1$ so this is

$1+ z+ z^2+z^3+ 5z^4+ 4+ 4z+ 4z^2+ 4z^3+ 5z^4= 5+ 5z^2+ 5z^3+ 5z^4= 5(1+ z+ z^2+ z^3+ z^4)$.

And, since $z^5= 1$, $z^5- 1= (z- 1)(z^4+ z^3+ z^2+ z+ 1)= 0$ and since $z- 1\ne 0$ for $z= e^{i\pi/5}$.....

No, the answer is NOT $
-5e^{\frac{3\pi\mathbf{i}}{5}}
$
Here is where the problem comes from: GRE Subject Tests: Mathematics

Number 43 of Math practice test.

6. Yes, and the correct answer is given there. But it isn't "E"!

7. Originally Posted by HallsofIvy
Yes, and the correct answer is given there. But it isn't "E"!
However, if you go to the answer key, it is E. Also, this was a test a question from 2005-2007 so if it isn't E, then the numerous amount of students who took the test then would have gotten 43 wrong if the answer is what you say. Due to the consequences of the test, I would think they would be extremely careful with developing the answer key so as not to hurt someones chances at getting accepted to a certain school by careless errors.

8. Originally Posted by dwsmith
However, if you go to the answer key, it is E.
The answer is indeed E. The test booklet is correct.
$1+ z+ z^2+z^3+ 5z^4+ 4z^5+ 4z^6+ 4z^7+4z^8+ 5z^9=$ $5(1+ z+ z^2+ z^3+ z^4)+5z^4$.
The key is in seeing it is to notice that $1+ z+ z^2+ z^3+ z^4 =0$
because $0=1-z^5=(1-z)( 1+ z+ z^2+ z^3+ z^4)$ and $z\not=1$

9. Originally Posted by Plato
The answer is indeed E. The test booklet is correct.
$1+ z+ z^2+z^3+ 5z^4+ 4z^5+ 4z^6+ 4z^7+4z^8+ 5z^9=$ $5(1+ z+ z^2+ z^3+ z^4)+5z^4$.
The key is in seeing it is to notice that $1+ z+ z^2+ z^3+ z^4 =0$
because $0=1-z^5=(1-z)( 1+ z+ z^2+ z^3+ z^4)$ and $z\not=1$
Why are tonio and HallsofIvy, who both do quality work in mathematics, so adamant it isn't?

10. Originally Posted by dwsmith
Why are tonio and HallsofIvy, who both do quality work in mathematics, so adamant it isn't?
I consider this to be a very tricky problem.
I also think 'copy & paste' has its own problems.
Look at
$1+ z+ z^2+z^3+ {\color{red}5z^4}+ 4+ 4z+ 4z^2+ 4z^3+ 5z^4= 5+ 5z^2+ 5z^3+$ $5z^4+{\color{red}5z^4}= 5(1+ z+ z^2+ z^3+ z^4)+{\color{red}5z^4}$.
I think a term was dropped.

11. Originally Posted by Plato
$1+ z+ z^2+z^3+ 5z^4+ 4z^5+ 4z^6+ 4z^7+4z^8+ 5z^9=$ $5(1+ z+ z^2+ z^3+ z^4)+5z^4$.
How were you able to do this factorization?

12. Originally Posted by dwsmith
How were you able to do this factorization?
That was done by HallsofIvy. He just dropped one term.

I might as well say this.
Once I saw that this was an official GRE publication, I knew the given answer had to be correct.
Nevertheless, I consider this to be a very poor question.
For one the given answer $- 5\exp \left( {\frac{{3\pi i}}{5}} \right)$ not in expected form.
I first thought this it was the obvious distracter because it is not usual to the minus there.
Only 26% got the question correct.

If $a=\exp \left( {\frac{{2\pi i}}{5}} \right)$ then $a^6=a,~a^7=a^2,~a^8=a^3,~\&~a^9=a^4$
I hope this helps.

13. Originally Posted by Plato
That was done by HallsofIvy. He just dropped one term.

I might as well say this.
Once I saw that this was an official GRE publication, I knew the given answer had to be correct.
Nevertheless, I consider this to be a very poor question.
For one the given answer $- 5\exp \left( {\frac{{3\pi i}}{5}} \right)$ not in expected form.
I first thought this it was the obvious distracter because it is not usual to the minus there.
Only 26% got the question correct.

If $a=\exp \left( {\frac{{2\pi i}}{5}} \right)$ then $a^6=a,~a^7=a^2,~a^8=a^3,~\&~a^9=a^4$
I hope this helps.
How are we transforming the constant of 1 to 5 WLOG and what happened to the $4z^5$?

14. Originally Posted by dwsmith
How are we transforming the constant of 1 to 5 WLOG and what happened to the $4z^5$?
Because $z^5=1$ we get $4z^5=4$.

15. Originally Posted by dwsmith
Why are tonio and HallsofIvy, who both do quality work in mathematics, so adamant it isn't?

I am not adamant at nothing: I just liked the trick Hallsofivy introduced in his response [taking into account that $z^5=1$ . Just as I didn't check thorughly my work I didn't check his either, and after Plato intervened I did check it. There was only one $5z^4$ lacking in Halls' work , and the way they put the answers in the GRE was very tricky ( I would even call it dirty ), no wonder Halls got confused:

$5z^4=5e^{8\pi i/4}=5e^{\pi i}e^{3\pi i/4}=-5e^{3\pi i/4}$

This is pretty easy, but when you have 170 minutes to solve 66 questions this can be way misleading...so misleading as to be considered unfair, perhaps.

I still have to be explained, and convinced, what good is to have this kind of exams with those times....what do they REALLY check and evaluate.

Tonio

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