Complex Numbers and Polynomials Problems

>>8488950
Bump, just post anything about complex numbers or polynomials. Just phrase it like a question.

>>8488950
>complex inequalities

complex number can't have an ordering

>>8488994
I know, we proved that aswell.

I meant stuff like lz - il < 1, involving the absolute value of complex numbers.

Please bring me problems man, I am starving. http://www.wtamu.edu/academic/anns/mps/math/mathlab/int_algebra/int_alg_tut25_poly.htm

Look at that shit. That is a university website and they ask problems like "what is the degree of this polynomial". I do not have an algebra book man, I need practice problems.

x^7=1
α ∉ R (alpha is a complex root of the equation)

Let θ = α + α^2 + α^4
Φ = α^3 + α^5 + α^6

Prove that Φ + θ = -1 and Φθ = 2

sorry I dont know what application people use to write the math symbols on /sci/ some1 please tell me

>>8489034
Latex.

>>8488950
>complex numbers and polynomials

Classes like that fell out of favor in the early 20th century. Most of those topics have been scattered among complex variable, number theory, and modern algebra books/classes.

Some of the chapters in this book deal with complex numbers at a similar level so try some of the exercises:
http://www.knowledge-dojo.com/papers/1930%20Advanced%20Trigonometry%20-%20Durell%20&%20Robson.pdf

>>8489034
Okay, I managed to figure out the first one.

First compute the polar form the "general" form of a and the notice that because the magnitude of a is 1, the magnitudes of all those powers of a is one.

Furthermore, by demoivres theorem we find that actually a^2. a^3, a^4, a^5 and a^6 are also roots of x^7. They are, actually, 6 distinct roots of x^7 = 1. And they are all complex.

Then we see that Φ + θ is the sum of 6 out of the 7 7-roots of 1. The only that is missing is the only real root of the equation, which is 1.

By another theorem, we know that the sum of every nth root of 1 equals 0 so we set

Φ + θ + 1 = 0 and then
Φ + θ = -1 follows directly from this.

Really interesting problem. Is there another way to address this problem? I think that by applying this last theorem I may be pretending the problem is easier than it really is. But I guess figuring out that I have to apply it is enough of a job.

I'll start working on the other proposition.

>>8489108
>>8489034

Okay, I got the second.

First, multiply all of that to get:
a^4 + a^6 + a^7 + a^5 + a^7 + a^8 + a^7 + a^9 + a^10

Then first remember that a^7 = 1
Then remember that a^8 = a, a^9 = a^2 and a^10 = a^7

So then the expression becomes

a^4 + a^6 + 1 + a^5 + 1 + a + 1 + a^2 + a ^3 =
a + a^2 + a^3 + a^4 + a^5 + a^6 + 3

But then we remember that this is actually equal to

Φ + θ + 3 =
-1 + 3 = 2

ez pz my man. Nice problem.

>>8489054
>Classes like that fell out of favor in the early 20th century. Most of those topics have been scattered among complex variable, number theory, and modern algebra books/classes.

I know. I don't have a class called Complex Numbers and Polynomials. I have this 2 semester course called 'Foundations'.

Last semester we did set theory and algebra (intro to groups, rings and fields). This semester we constructed the naturals from the Peano axioms, then the integers, then the rationals, then the reals, then the complex numbers and finally we studied polynomials.

So this course was literally about everything in mathematics, but it just so happens that tomorrows test is on the last 2 topics, complex numbers and polynomials.

And thanks for the reference.

>>8488994

However of course it is still sensible to speak of the inequality of complex numbers as-such; statements of the form z =/= w are of course sensible, on the contrary.

Bump.

I'm still alive here. Any more interesting problems?