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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?

 Dec 15, 2019
 #1
avatar+21725 
0

Isn't it just   4 + 5i ?  Complex roots come in pairs like   g +- k i

 

After a few calulations I find the other root to be

-6-3i

 Dec 15, 2019
edited by ElectricPavlov  Dec 15, 2019
edited by ElectricPavlov  Dec 16, 2019
 #2
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only if the coeffficients are real 

Guest Dec 15, 2019
 #4
avatar+108626 
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It states in that a and b are real.  The coefficients or z are clearly not real.

Melody  Dec 16, 2019
 #3
avatar+108626 
+1

Thanks EP.  

I want to give it a go myself.   laugh

 

z^2 + (a + 8i) z + (-39 + bi) = 0.

 

\([z-(4-5i)][z-(p+qi)]=z^2 + (a + 8i) z + (-39 + bi) \\ \text{Equating coefficients}\\ -(4-5i)-(p+qi)=a+8i\qquad and \qquad (4-5i)(p+qi)=-39+bi\\ -4+5i-p-qi=a+8i\qquad \qquad and \qquad 4p+5q+4qi-5pi=-39+bi\\ -4-p+(5-q)i=a+8i\qquad \qquad and \qquad 4p+5q+(4q-5p)i=-39+bi\\ -4-p=a \qquad 5-q=8\qquad \qquad and \qquad 4p+5q=-39 \qquad 4q-5p=b \\ -4-p=a \qquad \color{red}q=-3 \color{black}\qquad \qquad and \qquad 4p+5q=-39 \qquad 4q-5p=b \\ -4-p=a \qquad \qquad \qquad \qquad and \qquad 4p-15=-39 \qquad -12-5p=b \\ -4-p=a \qquad \qquad \qquad \qquad and \qquad \qquad \color{red}p=-6 \color{black}\qquad -12-5p=b \\ -4+6=a \qquad \qquad \qquad \qquad and \qquad \qquad \qquad \quad \color{black}\qquad -12+30=b \\ \color{red}a=2 \qquad \qquad\qquad \qquad \qquad and \qquad \qquad \qquad \quad \qquad\qquad b=18 \\\)

So we have

\([z-(4-5i)][z-(-6-3i)]=z^2 + (2 + 8i) z + (-39 + 18i) \\\)

 

I have not rechecked by expanding.     But I also get the other root to be -6-3i.

 

Is there an easier way to do it?

 

Coding:

[z-(4-5i)][z-(p+qi)]=z^2 + (a + 8i) z + (-39 + bi) \\
\text{Equating coefficients}\\

-(4-5i)-(p+qi)=a+8i\qquad and  \qquad (4-5i)(p+qi)=-39+bi\\
-4+5i-p-qi=a+8i\qquad \qquad and  \qquad 4p+5q+4qi-5pi=-39+bi\\

-4-p+(5-q)i=a+8i\qquad \qquad and  \qquad 4p+5q+(4q-5p)i=-39+bi\\
-4-p=a \qquad 5-q=8\qquad \qquad and  \qquad 4p+5q=-39 \qquad  4q-5p=b \\
-4-p=a \qquad \color{red}q=-3 \color{black}\qquad \qquad and  \qquad 4p+5q=-39 \qquad  4q-5p=b \\

-4-p=a \qquad \qquad \qquad \qquad and  \qquad 4p-15=-39 \qquad  -12-5p=b \\
-4-p=a \qquad \qquad \qquad \qquad and  \qquad \qquad \color{red}p=-6 \color{black}\qquad  -12-5p=b \\
-4+6=a \qquad \qquad \qquad \qquad and  \qquad \qquad \qquad \quad \color{black}\qquad  -12+30=b \\
\color{red}a=2 \qquad \qquad\qquad \qquad \qquad and  \qquad \qquad \qquad \quad \qquad\qquad  b=18 \\

 Dec 16, 2019
 #5
avatar+21725 
0

Hey Melody!  

  That is what I did (on paper) to find the other root.....I got the same numbers as you did for a and b.  I just didn't feel like transcribing it all on the keyboard when I was done ! cheeky   (Maybe for a REGISTERED user who wanted derivation I would have...)

 The question states a and b are real.....well , yes they are, but 'b' is part of   b i   (which obviously is not real)...a bit misleading, but at least we both figured it out....

ElectricPavlov  Dec 16, 2019
edited by ElectricPavlov  Dec 16, 2019

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