I have formed the following polynomial (from sum and product):


The zeros of this polynomial are ±2i, ±3i, 1±2i. I don't know how I can manipulate the polynomial to eliminate the undesired zeros.

  • 1
    $\begingroup$ With a complex zero, the conjugate is a zero too. $\endgroup$ – Wuestenfux Jun 7 '17 at 6:46
  • $\begingroup$ Think you have a sign error in the first factor, should be $= (z^2 + 4) (z^2 + 9) (z^2 \color{red}{+} 2 z + 5)$. $\endgroup$ – dxiv Jun 7 '17 at 6:51
  • $\begingroup$ I think the question is malformed because real polynomials always have complex conjugates as zeros - in other words, there is no way to eliminate the undesired conjugates. $\endgroup$ – Z. Aslam Jun 7 '17 at 6:58
  • $\begingroup$ get a REAL(ly complex) polynomial $\endgroup$ – Saketh Malyala Jun 7 '17 at 7:16

You can't get rid of the "extra" zeroes. If $f$ is a real polynomial, then $f(z)=0$ implies $f(\overline z)=\overline{f(z)}=0$.


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