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 $a=0$ .

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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 $P'(z)=z^2 + \frac{b}{a} z + \frac{c}{a} = 0$

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 $a=0$ .

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

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