66All the roots satisfy the given condition.
Find the number of roots of the equation z^7+4z^2+11=0 satisfying 1<|z|<2
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12 Answers
1sir are there 4 roots ?
since the sum of roots is 0
is one root is z then -z is another root
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also if z is one root then z is also a root and - z
so four roots are there
341I dont know how many distinct roots there are. There is atleast one real root (any odd degree polynomial has this propery).
But how on earth did u get that if z is a root, so is -z?
11Divide the given eqn on both sides by z^2 to have
z^5+4+\frac{11}{z^2}=0
Taking modulo on both sides we have \left|z^5+4 \right|=\frac{11}{|z^2|}\le |z|^5+4 from triangle's inequality, from which we get the obvious reasoning that |z|>1.
Similarly we have |z|^2=\left|\frac{11}{z^5+4} \right|\le \frac{11}{|z|^5+4}
From this we have that |z|<2....thus any z satisfying the given eqn must have its modulo between 1 and 2.
66@theprophet
now that you mentioned it, this is what Mathematica gives
Not much useful.
As far as I am concerned, I used the Rouche's theorem (which should be familiar if any one does a course on complex analysis) which gives easily the required number of zeroes in annulus 1<|z| <2.
341Rouche's theorem doesnt (directly) give the number of zeroes in an annulus.
You will first need soumik's result that we need |z|>1.
Soumik, your second inequality is incorrect. Would you like to try some more?
66More or less its direct by Rouche's.
For |z|=2,
|4z2+11| ≤ 4|z|2 + 11 =27 < 128 =|z7|
So in the disc |z|<2, the polynomial z7 + 4z2 +11 has the same number of roots as z7, namely, 7 (counting multiplicities).
On the other hand, on the unit circle |z|=1, the least value of |4z2+11| is 7 which is greater than 1=|z7|. As such on |z|=1,
|z7|<|4z2+11|
Hence, in the disc |z|<1, the polynomial z7 + 4z2 +11 has the same number of roots as 4z2+11, namely, 0.
And, obviously there are no roots with |z|=1.
Hence, all the roots of z7 + 4z2 +11 lie in the annulus 1 < |z| < 2.
11I know - I messed up the triangle's inequality....[4]
Dividing by z3 we have
z^4+\frac{4}{z}+\frac{11}{z^3}=0
Taking modulo on both sides we have |z|^4=\left|\frac{4}{z}+\frac{11}{z^3} \right|\le \frac{4}{|z|}+\frac{11}{|z|^3}
That gives |z|<2....
I think this is free from errors....
1ok but only location of roots is proved ...but we have to find no.of roots
66That's a 7th degree polynomial equation, so there are seven roots (counting multiplicities).