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Consider : $$P(z)= z^4 - 2z^3 + 6z^2 - 8z + 8$$ As the title says, find the roots of this complex quadratic equation having one purely imaginary root.

I need help with this problem, i am new with the ''complex world''. This is what i thought: Given a complex number: $z= a + bi$ where $a,b\in\mathbb C$, we know $P(z)$ has a pure imaginary root, then: $$P(bi)= (bi)^4 - 2(bi)^3 + 6(bi)^2 - 8(bi) + 8= 0$$ But i am stuck in here, i dont know how to proceed or if reasoning was correct. Any help would be helpful.

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  • $\begingroup$ The next step would be to use the definition of $i$. Do you need to find all roots or only the purely imaginary one? $\endgroup$ – klirk Mar 2 '18 at 17:48
  • $\begingroup$ I already know that my polynomial has a purely imaginary root. With that information i should be able to find all the other roots. The problem is i dont know how to proceed. $\endgroup$ – TheNicouU Mar 2 '18 at 17:49
  • $\begingroup$ This implies it’s got a factor of form $z^2+b^2$. Find the other one. $\endgroup$ – Macavity Mar 2 '18 at 17:51
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    $\begingroup$ that's not a quadratic equation $\endgroup$ – mercio Mar 2 '18 at 18:51
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    $\begingroup$ I would call it a quartic equation. Also, when I first read the question I thought you were saying the equation had exactly one imaginary root. (That's impossible, so presumably that's not what was meant.) It seems to me that a more natural phrasing would be "having a purely imaginary root." $\endgroup$ – David K Mar 2 '18 at 21:15
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Hint:

$i^2=-1$

This gives $$b^4+2b^3 i-6b^2-8bi+8=0 $$ Now use the fact that a complex number is zero if and only if its real and imaginary part are zero.

This will give you a system of two equations. Find the common solution to them.

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  • $\begingroup$ Extra hint: To solve the real half, let $x = b^2$, substitute, solve for $x$, and then reverse-substitute. The imaginary half is easier and does not require substitution. Also, bear in mind that both halves must produce a real solution because $b$ is a real number. $\endgroup$ – Kevin Mar 3 '18 at 1:05
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Since the coefficients of $P$ are purely real numbers, if $bi$ is a root of $P(z)$ then so is $\bar{bi}=-bi$. Hence, $(z-bi)(z+bi)=z^2+b^2\mid P(z)$. So $$P(z)=(z^2+b^2)(z^2+az+c)=z^4+az^3+(c+b^2)z^2+ab^2z+b^2c$$for some $a,c\in\Bbb C$. Compare coefficients with what we know $P(z)$ to be.

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  • $\begingroup$ We have $a,c\in \mathbb R$ $\endgroup$ – Macavity Mar 2 '18 at 18:03
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Hint:

If $ki$ is a solution also $-ki$ is a solution and the polynomial can be factorized as: $$(z^2+k^2)(az^2+bz+c)$$

can you find $a,b,c,k$?

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HINT...If $z=ai$ is a purely imaginary root, then so is $z=-ai$ since all the coefficients of the polynomial are real.

Therefore $(z^2+a^2)$ is a factor.

Therefore we can factorise the polynomial into the form $$(z^2+a^2)(z^2+bz+c)$$

it is then a simple matter to compare coefficients and find $a,b,c$

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