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Let a and b be real numbers. The complex number 4 - 5i is a root of the quadratic $z^2 + (a + 8i) z + (-39 + bi) = 0$. What is the other root?

I did a lot of work on hand and plugging this into the quadratic formula and with wolfram alpha but it doesn't simplify itself easily at all....anyone have any help to offer? Thanks!

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    $\begingroup$ how did you make use of given data that $4-5i $ is a root? $\endgroup$ – user87543 Jan 9 '14 at 17:29
  • $\begingroup$ Well I set the stuff plugged into the quadratic formula equal to it...but I don't see how I can proceed (not to mention there's a +/- sign and I don't know which one it is, positive or negative) $\endgroup$ – Freedom Jan 9 '14 at 17:30
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    $\begingroup$ If you have a $\pm$ sign, that means there are two solutions, one with $+$ and one with $-$. $\endgroup$ – Alex R. Jan 9 '14 at 17:34
  • $\begingroup$ Yeah, but how do I know whether 4-5i is the positive one or the negative one? $\endgroup$ – Freedom Jan 9 '14 at 17:36
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    $\begingroup$ On the complex plane, you don't have something like "positive" or "negative" - you can apply those terms only to the real and imaginary part of a complex number on their own, because $\Bbb{C}$ is not ordered. $\endgroup$ – Alex R. Jan 9 '14 at 17:42
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Since $$(4-5i)^2+(a+8i)(4-5i)+(-39+bi)=0$$$$\iff (4a-8)+(-5a+b-8)i=0,$$ we have $$4a-8=-5a+b-8=0\iff a=2,b=18.$$

Also, you'll have $$z^2+(2+8i)z+(-39+18i)=\{z-(4-5i)\}\{z-(-6-3i)\}.$$ So, what you want is $z=-6-3i$.

You can find this transformation by setting $\{z-(4-5i)\}\{z-(c+di)\}.$

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    $\begingroup$ You are welcome! My pleasure. $\endgroup$ – mathlove Jan 9 '14 at 17:38
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Let the other root be $u+iv$ where $u,v$ are real

Using Vieta's formula $\displaystyle u+iv+4-5i=-a-8i\iff -a-8i=u+4-i(5-v)$

Comparing the real & the imaginary parts,

$\displaystyle u+4=-a\iff u=-a-4$ and $5-v=8\iff v=-3$

Again, $\displaystyle(u+iv)(4-5i)=-39+bi\iff -(a+4+3i)(4-5i)=-(39-ib)$

Multiply & compare the real & the imaginary parts

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  • $\begingroup$ Thanks a lot that made it clearer xD $\endgroup$ – Freedom Jan 9 '14 at 17:38

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