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$$x^2 + 4x + 20 = 0$$ $$x=-2 \pm 4i$$

Roots of Quadtratic $z^2 + (4+i+qi)z + 20 = 0$ - Roots are $w$ and $w*$

(a) When $q$ is real, explain why it must be $-1$

Attempts:

Roots are conjugates therefore the coefficient must be real - and to get rid of the imaginary parts $q$ must be $-1$?

(b) Where $w=p+2i$ and $p$ is real, find the values of $q$.

Attempt:

Subbed $p+2i$ into equation to give $p=2$, where it is real. Substitute $2+2i$ into the equation again and get $q=3$, and $q=-8$? Very sure this is not correct - Should be a complex number?

Thank you.

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  • $\begingroup$ i think your first argument is ok $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '17 at 13:09
  • $\begingroup$ if for b) the same equation valid? $\endgroup$ – Dr. Sonnhard Graubner Apr 21 '17 at 13:12
  • $\begingroup$ This is all information given - I would assume so - I think possible roots of equations is required. $\endgroup$ – Cicada Apr 21 '17 at 13:17
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Hint for part (b): given the root $p+2i\,$, the other root is $\,\cfrac{20}{p+2i}\,$ by Vieta's relations.

Then, again by Vieta's relations for the sum of the roots:

$$ p+2i + \cfrac{20}{p+2i} = -(4+i+qi) $$

The latter gives $q$ in terms of $p$. The solution is not unique, for each $p \in \mathbb{R}$ there exists a complex $q$ given by the equation above such that $p+2i$ is a root.

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