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(a) Show that the roots of equation $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = 0$ where $a_1, a_2, a_3, a_4$ are real constants different from zero, has a pure imaginary root if $a_3^2 + a_1^2 a_4 = a_1 a_2a_3$
(b) Is converse of (a) true?

I tried factoring $z^4 + a_1 z^3 + a_2 z^2 + a_3 z + a_4 = (z^2 + b_1 z + b_2)(z^2 + c_1 z + c_2)$ and compared (equating both $b_1 = c_1 = 0$) but couldn't get the required results (that would get $a_1 = 0 $). Did I make any mistake on assumptions? I am beginning to doubt this question? Any suggestions, hints?

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When the roots are purely imaginary then $a_1$ is also purely imaginary and so cannot be "real constant different from zero". Something's wrong here... –  Marek Feb 27 '13 at 15:33
    
I don't know ... I am thinking the same. –  hasExams Feb 27 '13 at 15:34
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I bet the problem means that at least one of the roots is purely imaginary, not all of them. –  Thomas Andrews Feb 27 '13 at 15:39
    
uploaded, please check it :) –  hasExams Feb 27 '13 at 15:47
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The image you uploaded states "has a pure imaginary root", which is different than your stated question. I believe the question in the image meant "one of the roots has no real part". –  Jason Polak Feb 27 '13 at 16:03

2 Answers 2

up vote 5 down vote accepted

We will assume the problem means that the polynomial has a purely imaginary root, not that all the roots are purely imaginary.

The converse is not true unless you also know that $\frac{a_3}{a_1}>0$.

To do this, show that if $$a_3^2 + a_1^2 a_4 = a_1 a_2a_3$$ then $$z^4+a_1z^3+a_2z^2+a_3z+a_4 = (z^2+\frac{a_3}{a_1})(z^2+a_1z+\frac{a_4a_1}{a_3})$$

The coefficients for $x^3$ is $a_1$ on both side, the coefficient for $z$ is $a_3$ on both sides, and the constant is $a_4$ on both sides. So you only need to prove that the coefficient for $z^2$ is $a_2$ on both sides. On the right, that coefficient is $$\frac{a_4a_1}{a_3} + \frac{a_3}{a_1} = \frac{a_4a_1^2+ a_3^2}{a_1a_3}=\frac{a_1a_2a_3}{a_1a_3} = a_2$$

So the two polynomials are equal. But $z^2+\frac{a_3}{a_1}$ has a purely imaginary root when $\frac{a_3}{a_1}>0$, but does not when $\frac{a_3}{a_1}<0$. (The case of equal zero depends on purely imaginary, but since we know that $a_3\neq 0$, this is irrelevant.)

On the other hand, the other part $z^2+a_1z + \frac{a_4a_1}{a_3}$ never has purely imaginary roots when $a_1\neq 0$. So we can say that when we have real $a_1,a_2,a_3,a_4\neq 0$, then the polynomial $$z^4+a_1z^3+a_2z^2+a_3z+a_4$$ has a purely imaginary root if and only if $a_3$ and $a_1$ have the same sign and $a_3^2 + a_1^2 a_4 = a_1 a_2a_3$.

Note, the image from the book is actually the opposite problem, and it is an error on the part of the publisher/authors. It asks you to prove that the polynomial has a pure imaginary root if $a_3^2 + a_1^2 a_4 = a_1 a_2a_3$, which is not the case. For a simple example, let $(a_1,a_2,a_3,a_4)=(-1,1,1,-2)$. Then $$z^4+a_1z^3+a_2z^2+a_3z+a_4=(z-1)(z+1)(z^2-z+2)$$ which does not have a pure imaginary root.

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let me come again descriptively.

let $z=ix$ (purely imaginary)

After substituting the value of $z$ at given equation we get a polynomial in terms of $x$. (mark it 1).

Again $z=-ix$ (conjugate of $ix$) will also satisfy the polynomial. (mark it 2)

Subtract 1 from 2, and we'll get $x=\pm\sqrt{a_3/a_1}$. Use this value of $x$ in any of the equation (1 or 2). And we'll get the desired condition of having a purely imaginary root. (suggestion: have a pen and paper)

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