Question

If all zeros of a polynomial ()pz lies in a half plane then show that zeros of the derivatives '( ) pz also lie in the same half plane.

Answer 1

it is all constant SO derivative of constant is constant

Answer 2

Answer:

lf steady SO subordinate of consistent is steady

Explanation:

Show that the foundations of P' lie in similar half plane as the underlying foundations of P

Asked 6 years, 5 months prior

Altered 6 years, 5 months prior

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1

The issue proclamation is:

Part(a)

All expect P(z) is a non-steady polynomial with its foundations in some half plane H. All show that the subordinate P'(z) should likewise have its foundations in H.

Part (b)

Show that the underlying foundations of P' consistently exist in the shut arched body of the arrangement of foundations of P.

The Hint given is:

After clear decreases of the issue, check out at the way of behaving of P′P on the integral half-plane.

My considerations:

So that's what I believe, anything that half-plane H that we need to work with, we can continuously take a gander at P(eiθz), which has its zeros in the UHP, for a reasonable pivot by point theta.

Presently we can simply show that P′(eiθz)eiθ additionally has its underlying foundations in the UHP - - and afterward fixing the revolutions shows that both P and P' had its foundations in a similar half-plane H.

Utilizing the clue in regards to taking a gander at the reciprocal half-plane, we then, at that point, need to show that P' doesn't evaporate in the LHP (we definitely realize P doesn't disappear in that frame of mind, by supposition.)

Again following the clue we take a gander at P'/P in the LHP and show that it doesn't disappear. I'm stuck as of now. How might I continue?

Furthermore, with respect to part(b) what is a "raised structure"? I'm speculating that the solution to part (b) is quick, subsequent to sorting out part(a).