0568 #43

Forum for the GRE subject test in mathematics.
jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

0568 #43

Postby jayre » Wed Nov 04, 2009 10:43 pm

I'm banging my head against the wall on this one.

43. if z = exp(2*Pi*i/5), then 1 + z + z^2 + z^3 + 5z^4 + 4z^5 + 4z^6 + 4z^7 + 4z^8 + 5z^9 = ...

The answer is -5*exp(3*Pi*i/5), and I can't seem to get there.

I haven't had complex analysis, so I only know what I've picked up from using it in other classes. I can get the given expression to equal:

5(1 + exp(Pi*i/5) + exp(2*Pi*i/5) + 2*exp(3*Pi*i/5) + exp(4*Pi*i/5))

but I have no idea how to combine those exponentials. I know they're all 5th roots of unity, but I don't see how that helps. Any ideas?

(Also, sorry, I've never used LaTex, so this looks crappy. I'll learn it when I need it.)

mag487
Posts: 13
Joined: Thu Aug 06, 2009 3:59 am

Re: 0568 #43

Postby mag487 » Wed Nov 04, 2009 11:22 pm

If z = e^(2*i*pi/5), then z^5 = 1 and z^i = z^(i+5). So we can express the sum

1 + z + z^2 + z^3 + 5z^4 + 4z^5 + 4z^6 + 4z^7 + 4z^8 + 5z^9

as

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

Now, (1 + x + x^2 + ... + x^n) = (1 - x^(n+1))/(1 - x). So if x^(k+1) = 1 and x doesn't equal 1, it follows that 1 + x + x^2 + ... + x^k = 0. Therefore 1 + z + z^2 + z^3 + z^4 = 0, and the rewritten sum in question is 5*0 + 5z^4 = 5e^(8*i*pi/5) = -5e^(3*i*pi/5).

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: 0568 #43

Postby joey » Wed Nov 04, 2009 11:44 pm

We can also use the sum-of-roots theorem. The fifth roots of unity are all the zeros of the polynomial
z^5-1=0,
so they must sum to -\frac{0}{1}=0.

Sum-of-roots seems like a good one to know for the exam. I got this problem wrong when I took the practice, too, so now I have flash cards for sum-of-roots and product-of-roots. Rational roots is handy, too.
Last edited by joey on Wed Nov 04, 2009 11:53 pm, edited 1 time in total.

jayre
Posts: 11
Joined: Wed Nov 04, 2009 10:30 pm

Re: 0568 #43

Postby jayre » Wed Nov 04, 2009 11:50 pm

Thanks for your quick replies, guys. Both methods make perfect sense, and both are things I just wouldn't have thought of even trying... I guess it's also better to leave everything in terms of z rather than writing them all out as exponentials. I'll have to get more comfortable with complex numbers by Saturday :-p

joey
Posts: 32
Joined: Fri Oct 16, 2009 3:53 pm

Re: 0568 #43

Postby joey » Thu Nov 05, 2009 8:05 am

No problem. This forum is great because it gets us all thinking about the appropriate topics in a way that helps one another do better. It's one of few ways in which I use the internet for its original purpose: academic collaboration over vast distances. (At least, I assume you're not all concentrated in the Twin Cities metro.)

When you study complex, be sure to learn the Cauchy-Riemann equations and the Cauchy-Goursat theorem (sometimes called the Cauchy integral theorem). These are favorites on past GREs, and they're easily accessible. If you have more time and really want to dig in, learn about Laurent series / residues. These topics require you to think about series expansion in a more rigorous way, and they sometimes appear on the GRE.

Good luck!




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