My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree3Thanks
  • 1 Post By CaptainBlack
  • 1 Post By CaptainBlack
  • 1 Post By CaptainBlack
Reply
 
LinkBack Thread Tools Display Modes
October 31st, 2014, 01:39 PM   #1
Senior Member
 
Joined: Oct 2014
From: Complex Field

Posts: 119
Thanks: 4

Complex polynomial and Vieta's formulas

Hello,

I had a question in my homework that I managed to solve part A of it with one of the Vieta's formulas, but part B was wrong. I am almost sure that part B also uses one of the Vieta's formulas but I just don't know how to use it properly. The question is:

P(z)=z^7 + a6*z^6 + a5*z^5 + a4*z^4 + a3*z^3 + a2*z^2 +a1*z + a0

provided information:

z1=1+2i is the root of P(z),P'(z) and P''(z) ;
a0=-53

A) Real root of P(z) is _____ ?
B) a6=?

My answer to A was :
from the information we can say that (1+2i) is root of multiplicity 3. and because it's a complex number also the conjugate is of multiplicity 3.
and the only one left is a real (because of 7 roots) so I mark it as 't' here.

So I used Vieta's formula for multiplying and got:
(1+2i)^3*(1-2i)^3*t = (-1)^7*(a0/a7) => // a0=-53 ; a7=1
=> 125*t = (-1)*(-53/1)
=>t = 53/125

This is part A and is correct.

Now what I tried to do for part B was to use the Vieta's sum formula in this way:

The sum of complex roots is 0, and the remaining root is 53/125.
So:

0+53/125 = -a6/a7 => // a7=1
a6=-53/125 < But that's a wrong answer.

What am I doing wrong?

Thanks!
noobinmath is offline  
 
October 31st, 2014, 02:20 PM   #2
Senior Member
 
Joined: Jan 2012
From: Erewhon

Posts: 245
Thanks: 112

Quote:
Originally Posted by noobinmath View Post
Hello,

I had a question in my homework that I managed to solve part A of it with one of the Vieta's formulas, but part B was wrong. I am almost sure that part B also uses one of the Vieta's formulas but I just don't know how to use it properly. The question is:

P(z)=z^7 + a6*z^6 + a5*z^5 + a4*z^4 + a3*z^3 + a2*z^2 +a1*z + a0

provided information:

z1=1+2i is the root of P(z),P'(z) and P''(z) ;
a0=-53

A) Real root of P(z) is _____ ?
B) a6=?

My answer to A was :
from the information we can say that (1+2i) is root of multiplicity 3. and because it's a complex number also the conjugate is of multiplicity 3.
and the only one left is a real (because of 7 roots) so I mark it as 't' here.

So I used Vieta's formula for multiplying and got:
(1+2i)^3*(1-2i)^3*t = (-1)^7*(a0/a7) => // a0=-53 ; a7=1
=> 125*t = (-1)*(-53/1)
=>t = 53/125

This is part A and is correct.

Now what I tried to do for part B was to use the Vieta's sum formula in this way:

The sum of complex roots is 0, and the remaining root is 53/125.
So:

0+53/125 = -a6/a7 => // a7=1
a6=-53/125 < But that's a wrong answer.

What am I doing wrong?

Thanks!
The sum of the complex roots is not zero, the imaginary part of the sum of the complex roots is zero.

Also from what you have written it is evident that you think you are dealing with a real polynomial, but have nowhere stated it.

CB
Thanks from noobinmath
CaptainBlack is offline  
October 31st, 2014, 02:33 PM   #3
Senior Member
 
Joined: Oct 2014
From: Complex Field

Posts: 119
Thanks: 4

Thanks,

But as for part A, did I do something wrong and by chance the answer is correct? Or part A is alright? And also give me a tip for part B?

Last edited by noobinmath; October 31st, 2014 at 02:36 PM.
noobinmath is offline  
October 31st, 2014, 03:06 PM   #4
Senior Member
 
Joined: Jan 2012
From: Erewhon

Posts: 245
Thanks: 112

Quote:
Originally Posted by noobinmath View Post
Thanks,

But as for part A, did I do something wrong and by chance the answer is correct? Or part A is alright? And also give me a tip for part B?
If $P(x)$ is a real polynomial then your approach to part A is OK.

For part B the sum of the roots is 6+53/125 (assuming you answer to part A is numerically correct). This is because each of the complex roots has real part 1 and so the real part of the sum of roots is 6+53/125 and the imaginary parts cancel as the complex roots occur in conjugate pairs.

CB
Thanks from noobinmath
CaptainBlack is offline  
October 31st, 2014, 03:12 PM   #5
Senior Member
 
Joined: Oct 2014
From: Complex Field

Posts: 119
Thanks: 4

Thank you, the first part was numerically correct. But can you tell me what is the real approach for it? Because I want to get to the right answer in the right way, and not by a chance. And also, how come that I did it in the real polynomial approach and still got the right answer? (How would you do that?)

Last edited by noobinmath; October 31st, 2014 at 03:19 PM.
noobinmath is offline  
October 31st, 2014, 11:20 PM   #6
Senior Member
 
Joined: Jan 2012
From: Erewhon

Posts: 245
Thanks: 112

Quote:
Originally Posted by noobinmath View Post
Thank you, the first part was numerically correct. But can you tell me what is the real approach for it? Because I want to get to the right answer in the right way, and not by a chance. And also, how come that I did it in the real polynomial approach and still got the right answer? (How would you do that?)
There is nothing wrong with what you did, either the question as asked states the a's are real, implicitly implies it or it can be proven (or something equivalent can be proven).

What we know is that the product of the roots is real the product of the four mystery roots is a real multiple of $(1-2i)^3$ and that is all we can say which is insufficient to fix even one real root. For example the following polynomial satisfies everything given but does not have real coefficients other than the first and last:
$$
P(x)=\left( x-1\right) \,{\left( x-\frac{\sqrt{53}}{{5}^{\frac{3}{2}}}\right) }^{2}\,{\left( x-2\,i-1\right) }^{3}\,\left( x-2\,i+11\right) $$
and it has a real root different from the "correct answer"

What I am trying to say is that there is insufficient information to solve the asked questions A and B without the a's being real.

CB
Thanks from noobinmath

Last edited by CaptainBlack; October 31st, 2014 at 11:23 PM.
CaptainBlack is offline  
November 1st, 2014, 01:29 AM   #7
Senior Member
 
Joined: Oct 2014
From: Complex Field

Posts: 119
Thanks: 4

Oh so the polynomial is considered "real" not because of "z" but because of the a's? So yes, the a's are real!, sorry for forgetting to mention that.
And now B part is also correct, thank you! (It is 6+53/125)
noobinmath is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
complex, formulas, polynomial, vieta



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Complex polynomial muhsentdrawjiac Algebra 5 July 12th, 2012 12:41 AM
absolute value of a complex polynomial Broodjekaas Complex Analysis 1 June 9th, 2011 06:30 AM
integral of a complex polynomial Broodjekaas Complex Analysis 0 June 8th, 2011 01:14 AM
complex polynomial Vygotsk Complex Analysis 2 May 17th, 2010 03:31 PM
complex polynomial ^e^ Complex Analysis 1 April 1st, 2007 04:45 AM





Copyright © 2019 My Math Forum. All rights reserved.