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# Test of Mathematics Solution Subjective 87 – Complex Roots of a Real Polynomial

In Progress 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.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

## 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 $$a=0$$.