how many zeros polynomial P(z)= 3z^6 +z^5+ z^4+ 2z^3 + 10z^2+ z+1 has in area 1

Question

how many zeros polynomial  P(z)= 3z^6 +z^5+ z^4+ 2z^3 + 10z^2+ z+1 has in area 1<|z|<2?

anonymous · Answer

lalay is typing a reply..

anonymous · Answer

tell me how to begin this problem :P

gg · Answer

I have to see step by step of solving, I can't just write a number of zeros on exam :D

lalaly · Answer

lol wait im thinkin

anonymous · Answer

lalaly is now thinking.. please wait

gg · Answer

hahahah :D ok :D you are lalaly's lawer? :D

lalaly · Answer

i tried but i got the answer wrong...im sorryyy

anonymous · Answer

how to start.. tell me how to start :'(

gg · Answer

I don't know :D If I know how to start, I;ll probably know how to finish :D I posted another one problem :) are u maybe good in probabillity?

zarkon · Answer

there are 4, but I used my calculator to get that answer

anonymous · Answer

hold on

anonymous · Answer

by the looks of this question, although there is no indication given, i think this is a polynomial of a complex variable and 
1<|z|<2 means the annulus  
i think

zarkon · Answer

all 6 roots are complex

anonymous · Answer

oh ok.  isn't there a formula that counts the number of zeros minus the number of poles?

anonymous · Answer

and since this obviously has no poles it is just he number of zeros. you integrate p'(x)/p(x) maybe

zarkon · Answer

I am unfamiliar with that

anonymous · Answer

i think so it is like residue theorem.  divide by 2 pi i too.

zarkon · Answer

rouche's theorem

anonymous · Answer

problem is i forget how to integrate this thing around the annulus!

anonymous · Answer

maybe you are not supposed to integrate. maybe you are supposed to bound then?

anonymous · Answer

god i have forgotten everything. lucky i remember my phone number

anonymous · Answer

i am going to try this and see if it works. show that all 6 zeros are inside |z|<2 and then show 2 are inside |z|<1 but i may not be able to

my guess is that is the method though.  that or integrating 

$$\frac{1}{2\pi i}\int_{1<|z|<2} \frac{p'(z)}{p(z)}$$

anonymous · Answer

couldn't make first method work.  help!!

zarkon · Answer

ok...
if |g(z)|<|f(z)| for all z on gamma(gamma a simple loop), then f and f+g have the same number of roots inside gamma

let $$f(z)=3z^6$$

and $$g(z)=z^5+ z^4+ 2z^3 + 10z^2+ z+1$$

for $$|Z|=2$$

$$|f(2)|=3(2)^6=192$$

and $$|g(2)|=2^5+ 2^4+ 2\cdot 2^3 + 10\cdot 2^2+ 2+1=106$$

thus $$p(z)$$has the same number of roots as $$3z^6$$ which is 6


not let $$f(z)=10z^2$$ and $$g(z)=3z^6+z^5+ z^4+ 2z^3 + z+1$$

then for |Z|=1

|f(1)|=10 and |g(1)|=3+1+1+2+1+1=9

$$10z^2$$ has 2 roots in|Z|<1 thus 
P(Z) also has 2 roots

hence

P(z) has 6-2=4 roots in 1<|Z|<2

anonymous · Answer

oh very nice!  i got the first part with the 2 (by copying right off of wiki) but didn't get the second part with the 10z^2.

zarkon · Answer

well I learned something new ;)

anonymous · Answer

i guess i could have could have thought somewhat harder about the 10z^2 part since i wanted to show that there were TWO zeros inside |z|<1 !  nice work

anonymous · Answer

i learned something new as well.  happens every once in a while

zarkon · Answer

:)

anonymous · Answer

hope Gg comes back and they can learn something as well

zarkon · Answer

yes

gg · Answer

thank you very much :) but how do I know what tochoose for f(z) and what for g(z)?