# Find the real, monic polynomial of the lowest possible zeros which has zeros -1-2i, -3i and 2i.

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

$$p(z)=(z^2-2z+5)(z^2+9)(z^2+4)$$

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.

• With a complex zero, the conjugate is a zero too. – Wuestenfux Jun 7 '17 at 6:46
• Think you have a sign error in the first factor, should be $= (z^2 + 4) (z^2 + 9) (z^2 \color{red}{+} 2 z + 5)$. – dxiv Jun 7 '17 at 6:51
• 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. – Z. Aslam Jun 7 '17 at 6:58
• get a REAL(ly complex) polynomial – 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 $f(\overline z)=\overline{f(z)}=0$.