# GRE Math Subject Practice Test #43

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1. Sep 14, 2008

### Zabopper

First of all, I hope I'm posting this in the appropriate forum and let me know if I'm not.

I did pretty okay on this practice test, but I never took complex analysis, though I've tried to teach myself the rudiments. Maybe that's not even the problem, but this seems kind of basic, something I shouldn't be missing. So I'm looking for help with how to solve this problem, and also suggestions for preparing for the exam so that I don't miss this category of problem come test day. Here it is:

If z = e^(2*pi *i / 5), then 1 + z + z^2 + z^3 +5*z^4 + 4*z^6 + 4*z^7 +4*z^8 +5*z^9 =

a)0

b) 4*e^(3*pi*i / 5)

c) 5*e^(4*pi*i / 5)

d) -4*e^(2*pi*i / 5)

e) -5*e^(3*pi*i / 5)

Thank you everyone!

2. Sep 15, 2008

### sutupidmath

$$e^{xi}=cosx+isinx$$ NOw u have:

$$z= e^{{2\pi}\frac{i}{5}}=cos\frac{2\pi}{5} +isin\frac{2\pi}{5}$$

$$z^2=e^{2*{2\pi}\frac{i}{5}}=cos\frac{4\pi}{5}+isin\frac{4\pi}{5}$$

Now u can play a lill bit with trig functions of a double angle, so some things will cancel out. Also

$$1=e^{{2\pi}i}=cos2\pi+isin{2\pi}$$

i think that doing this for the whole powers of z and looking for a pattern of how things will cancle out, you should be able to get to the result.

3. Sep 15, 2008

### sutupidmath

There might be shortcuts though, but none of which i can think at the moment!

4. Sep 15, 2008

### HallsofIvy

Staff Emeritus
I think that's the hard way to do it. It is much simpler to do the multiplications in exponential form than "cos + i sin".

If $$z= e^{2\pi i/5}[/itex] then [tex]z^2= e^{4\pi i/5}$$
$$z^3= e^{6 pi i/5}= e^{pi i}e^{\pi i/5}= -e^{\pi i/5}$$
$$z^4= e^{8\pi i/5}= -e^{3\pi/5}$$
$$z^5= e^{10\pi i/5}= e^{2\pi i}= 1$$
$$z^6= e^{12\pi i/5}= e^{2\pi i}e^{2\pi i/5}= e^{2\pi i/5}$$
$$z^7= e^{14\pi i/5}= e^{2\pi i}e^{4\pi i/5}$$
$$z^8= e^{16\pi i/5}= e^{3\pi i}e^{\pi i/5}= -e^{\pi i/5}$$
and
$$z^9= e^{18\pi i/5}= e^{3\pi i}e^{3\pi i/5}= -e^{3\pi i/5}$$

You should be able to put those into your formula and come up with an answer.