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# Complex Roots of a Real Polynomial (TOMATO Subjective 87)

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.

October 24, 2016