Test of Mathematics Solution Subjective 87 - Complex Roots of a Real Polynomial

This is a Test of Mathematics Solution Subjective 87 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

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Problem:

Let \(P(z) = az^2+ bz+c\), where \(a,b,c\) are complex numbers.

\((a)\) If \(P(z)\) is real for all real \(z\), show that \(a,b,c\) are real numbers.

\((b)\) In addition to \((a)\) above, assume that \(P(z)\) is not real whenever \(z\) is not real. Show that \(a=0\).

Solution:

\((a)\) As \(P(z)\) is real for all real \(z\), we have \(P(0)=c\) \(=> c\) is real.

\(P(1) = a+b+c\) is real.

\(P(-1) = a-b+c\) is real.

\(P(1) + P(-1) = 2a+2c\) is real.

As \(c\) is real \(=> a\) is also real.

Similarly as \((a+b+c)\) is real \(=> (a+b+c)-(a+c)\) is also real.

Implying \(b\) is also real.

Thus all \(a,b,c\) are real.

\((b)\)Let us assume that \(a\neq 0\).

Thus the equation can be written as \(P'(z)=z^2 + \frac{b}{a} z + \frac{c}{a} = 0\)

Let \(\alpha\) be a root of the equation. If \(\alpha\) is imaginary that means \(P'(\alpha)\) is imaginary. But \(P'(\alpha)=0\), thus \(\alpha\) is real. Similarly \(\beta\), the other root of the equation, is also real.

Therefore \(\alpha + \beta = -\frac{b}{a}\). \(\cdots (i)\)

Take \(x=\frac{\alpha + \beta}{2} + i\)

Then \(P'(x) = \frac{(\alpha + \beta)^2}{4} + (\alpha + \beta)i -1 + \frac{b}{a}(\frac{\alpha + \beta}{2} + i) + \frac{c}{a}\)

\(=> P'(x) = \frac{(\alpha + \beta)^2}{4} + (\alpha + \beta)i -1 + \frac{b}{a} \frac{\alpha + \beta}{2} +\frac{b}{a} i + \frac{c}{a}\)

Using \((i)\), we get,

\(=> P'(x) = \frac{(\alpha + \beta)^2}{4} - \frac{b}{a} i -1 + \frac{b}{a} \frac{\alpha + \beta}{2} +\frac{b}{a} i + \frac{c}{a}\)

\(=> P'(x) = \frac{(\alpha + \beta)^2}{4} -1 + \frac{b}{a} \frac{\alpha + \beta}{2} + \frac{c}{a}\)

Thus \(P'(x)\) is real even when \(x\) is imaginary. Thus our assumption that \( a \neq 0\) is wrong.

Hence Proved \(a=0\).

Some Direct Inequalities | TOMATO Subjective 80

This is a beautiful problem based on Some Direct Inequalities from Test of Mathematics Subjective Problem no. 80.

Problem: Some Direct Inequalities

If \(a,b,c\) are positive numbers, then show that

\(\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\geq a+b+c\)

Solution: This problem can be solved using a direct application of the Titu's Lemma but we will instead prove the lemma first using the Cauchy-Schwarz inequality.

According to the Cauchy-Schwarz inequality we have,

\(\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)\ge \left(a_1b_1+a_2b_2+\cdots+a_nb_n\right)^2 \)

Replacing \(a_i\to \dfrac{a_i}{\sqrt{b_i}}\) and \(b_i\to \sqrt{b_i}\) we get,

\(\left(\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots +\dfrac{a_n^2}{b_n}\right)\left(b_1+b_2+\cdots+b_n\right)\ge \left(a_1+a_2+\cdots+a_n\right)^2,\)

which is equivalent to

\(\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots+\dfrac{a_n^2}{b_n}\geq \dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{b_1+b_2+\cdots+b_n}\)

Now this inequality is referred to as the Titu's Lemma.

This brings us to the problem which can be observed to be a simple application of the lemma. We just need to make the following substitutions.

\(a_1=b, a_2=c\) and \(b_1=b_2=b+c\)

Then we have,

\(\dfrac{b^2}{b+c}+\dfrac{c^2}{b+c}\geq \dfrac{(b+c)^2}{2(b+c)}\)

\(=>\dfrac{b^2+c^2}{b+c}\geq \dfrac{b+c}{2}\)

Thus similarly we have,

\(=>\dfrac{c^2+a^2}{c+a}\geq \dfrac{c+a}{2}\) and \(=>\dfrac{a^2+b^2}{a+b}\geq \dfrac{a+b}{2}\)

Adding the three inequalities we get,

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq \dfrac{b+c}{2}+\dfrac{c+a}{2}+\dfrac{a+b}{2}\)

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq \dfrac{2(a+b+c)}{2}\)

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq {a+b+c}\)

Hence Proved.

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