# Search results

1. ### Finding Complex Roots

Ah yes so forget the negative ninth pi, was 17pi/9 correct? And 19pi/9?
2. ### Finding Complex Roots

Cosine is an even function so throw in -pi/9 as well. As for the third how about 17∏/9?
3. ### Finding Complex Roots

cos(pi/3) = cos(3θ) cant I take cos inverse of both sides yielding pi/3 = 3θ? then θ=1/9 pi of course times some scalar k?
4. ### Finding Complex Roots

Sure so r = 1 and θ=1pi/9
5. ### Finding Complex Roots

Ok we have cosine of an angle corresponding to 1/2 and the sin corresponding to (sqrt3)/2 so the θ in question should be ∏/3 (60 degrees). The r = (1/2)^2 + (sqrt3/4)^2 r = 1/4 + 3/4 r = 1
6. ### Finding Complex Roots

That kind of looks like the set up to DeMoivre theorem? I know if it is r(cosθ + isinθ)^n = r^n(cos nθ + isin nθ) But how does this help us?
7. ### Finding Complex Roots

No my book never mentioned anything about this, but it makes sense that we have a positive and a negative version of the root. That would give me four if I included the conjugates to the one I wrote, what would the 3rd set be??
8. ### Finding Complex Roots

Oh no, do you get something different when you take the cube root?
9. ### Finding Complex Roots

However we would have 6 roots to this polynomial would we not, by the fundamental theorem of algebra?
10. ### Finding Complex Roots

OMG how silly of me yes you'd take the cube root of the positive and negative answers each so, z = (-1 + i3^1/6)/(2^1/3) z = (-1 - i3^1/6)/(2^1/3)
11. ### Finding Complex Roots

Wicked alright let u = z^3 We have u^2 + u + 1 = 0 then u = -1/2 +- [(√3)/2]i Now given that we took u = z^3, can I just cube the results to get the answers for z??
12. ### Finding Complex Roots

Homework Statement Find all complex solutions of z^6 + z^3 + 1 (z^3 + 1)/(z^3 - 1) = i Homework Equations The Attempt at a Solution I am going crazy with trial and error with these, there must be some systematic method or tricks that I am oblivious of. For the second question I...
13. ### Cardinality of a subset of [0,1]

Hmm, ok well could the sequence be 0.1, 0.101, 0.010101... till I hit that n (n being the finite number of non zero digits?
14. ### Cardinality of a subset of [0,1]

It seems so, but the unit interval has cardinality c and at first glance it appeared to me that we were excluding a finite set from an infinite one, and if I am not mistaken with cardinal arithmetic, c would remain the cardinality. But it seems to be otherwise now.
15. ### Cardinality of a subset of [0,1]

Could I go, f(1) = 0.11 f(2) = 0.12 f(3) = 0.13 ... and skip numbers like 0.2, 0.3? But then what happens to all the numbers in between I'm missing?
16. ### Cardinality of a subset of [0,1]

Homework Statement What is the cardinality of the set of all numbers in the interval [0, 1] which have decimal expansions with a finite number of non-zero digits? Homework Equations The Attempt at a Solution I say its still c? Am I correct, there is no way I can pair this set with the...
17. ### Infinite/Finite Set Proof

I guess how I would write it is problematic for me, I understand your idea, take that excluding set A, which is a set from x1 to xn and begin the mapping from a xn + 1. But how do I know my excluded set is ordered, and how do I know that the numbers are consecutively equally spaced the set from...
18. ### Infinite/Finite Set Proof

Sure I'd map 1 from the first set N to say 11, map 2 to 12, 3 to 13.... My bijection would be for n in N (the fully inclusive N) f(n) = n+ 10?
19. ### Infinite/Finite Set Proof

I know to show cardinalities are equal I need to be able to construct a bijection between them... Bijection between two infinite sets seems confusing.
20. ### Infinite/Finite Set Proof

Homework Statement Let S be in finite and, A a subset of S be finite. Prove that that the cardinality of S = the cardinality of S excluding the subset of A. Homework Equations The Attempt at a Solution We can write out the finite subset A as (x1, x2, ... xn) which can be put...
21. ### Cardinality of the set of all functions

Interesting well I know the set N has 2^(aleph nought) subsets and the set S = {1, 2} has 2^2 subsets, so am I correct when I say: 2^(aleph nought) x 2^2 = 2^(aleph nought + 2)? Also I am trying to more so understand your explanation of the equivalence of a function from A to B to a subset...
22. ### Cardinality of the set of all functions

Homework Statement What is the cardinality of the set of all functions from N to {1,2}? Homework Equations The Attempt at a Solution I know the cardinality of the set of all functions coincides with the respective power set (I think) so 2^n where n is the size of the set. The...
23. ### RSA Encryption

Homework Statement "You are to receive a message using the RSA system. You choose p = 5, q = 7 and E = 5. Verify that D = 5 is a decoder. The encoded message you receive is 17. What is the actual (decoded) message?" Homework Equations The Attempt at a Solution N= pq = 35, our E...

25. ### Fermat Little Theorem?

Haha why am I even asking, of course I can... I think its getting too late for me! I will just simplify 10^6 - 1 into (10^3 + 1)(10^3 - 1) and following by the same algorithm, 3 will be my appropriate divisor and thus final prime factor. Thanks Micromass. EDIT: Wait I made a careless error...
26. ### Fermat Little Theorem?

Oh wow, yes I can now write 10^6 - 1 using FLT again and it becomes 10^6 is congruent to 1(MOD7)! So now I have 2 primes that divide according to the condition, 7 and 13. However is the third prime found by a similar technique?
27. ### Fermat Little Theorem?

Interesting... So we could have (10^6 +1)(10^6 - 1) is congruent to 1(mod13). Can I now use this theorem? "If p is a prime number and x is an integer satisfying x^2 - 1 (mod p), then either x  is congruent to 1 (mod p) or x is congruent to p-1 (mod p)." But does it even help?
28. ### Fermat Little Theorem?

Homework Statement Find 3 different prime factors of 10^12 -1. Homework Equations The Attempt at a Solution I began trying to solve this with help from F.L.T. - if p is prime and p does NOT divide a, then a^(p-1) is congruent to 1 (mod p). So I re-wrote, 10^(13-1) is...
29. ### Wilson's Theorem

Oh I see... thanks very much. That indeed demonstrates there to be more than one possible solution for x.
30. ### Wilson's Theorem

Hmm, well I can see that they are all 17 apart (17 like the modulus), is there a more mathematical way to express or suggest that?