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Let $b$ and $c$ be complex constants such that $z^2$ + $bz$ + $c$ = $0$ has two different real roots. Show $b$ and $c$ are real.

I think I need to be using Vieta's formula, however I have solved it a different way and I wasn't sure if my answer is valid or if I should rework this answer somehow?

Let $z_1$, $z_2$ be roots $\in$ $\Re.$ Then we have $z_1^2+bz_1+c=0$ and $z_2^2+bz_2+c=0.$ Since $b$ and $c$ are complex constants we know that $b=b_1+b_2i$ and $c+c_1+c_2i.$ Now we can plug that information into $z_1^2+bz+c=0$ and $z_2^2+bz+c=0.$ $$\Downarrow$$ $$z_1^2+(b_1+ib_2)z_1+(c_1+ic_2)=0$$ $$\Downarrow$$ $$z_1^2+z_1b_1+c_1+i(b_2z_1+c_2)=0$$ $$and$$ $$z_2^2+(b_1+ib_2)z_2+(c_1+ic_2)=0$$ $$\Downarrow$$ $$z_2^2+z_2b_1+c_1+i(b_2z_2+c_2)=0.$$

We also know that $z_1$ $\neq$ $z_2$ and using this information we have $z_1^2+z_1b_1+c_1+i(b_2z_1+c_2)=0$ and $z_2^2+z_2b_1+c_1+i(b_2z_2+c_2)=0$. Now, by taking the imaginary parts, we have: $$i(z_1b_2+c_2)=0$$ $$and$$ $$i(z_2b_2+c_2)=0.$$ Now we can set these two equations equal to each other and solve for a real solution.$$i(z_1b_2+c_2)=i(z_2b_2+c_2)$$ $$\Downarrow$$ $$iz_1b_2+ic_2=iz_2b_2+ic_2$$ $$\Downarrow$$ $$iz_1b_2+ic_2-ic_2=iz_2b_2+ic_2-ic_2$$ $$\Downarrow$$ $$iz_1b_2=iz_2b_2$$ $$\Downarrow$$ $$\frac{iz_1b_2}{ib_2}=\frac{iz_2b_2}{ib_2}$$ $$\Downarrow$$ $$z_1=z_2,$$ which is a contradiction, thus we see for this to be true $b_2$ = $c_2$ = $0$. Therefore, $b_2$ = $c_2$ = $0$, where both $b$ and $c$ are real.

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  • $\begingroup$ $z^2 + b z + c = (z-r_1)(z-r_2) = z^2 + (-r_1 - r_2)z + r_1 r_2$ $\endgroup$ – Will Jagy Jul 18 '14 at 16:58
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You can to that way, but you're doing too much work.

Here's a simpler version of your computations: write $b=b_1+ib_2$, $c=c_1+ic_2$; then you have $$ z^2+b_1z+c_1+i(b_2z+c_2)=0 $$ Let $r$ and $s$ be the two roots: then $$ r^2+b_1r+c_1+i(b_2r+c_2)=0 \qquad\text{and}\qquad s^2+b_1s+c_1+i(b_2s+c_2)=0 $$ so $$ r^2+b_1r+c_1=0 \qquad\text{and}\qquad s^2+b_1s+c_1=0 $$ for the real part; next $$ b_2r+c_2=0 \qquad\text{and}\qquad b_2s+c_2=0 $$ for the imaginary parts. Therefore $-c_2=b_2r=b_2s$, from which $$ b_2(r-s)=0. $$ Owing to $r\ne s$, we get $b_2=0$ and hence also $c_2=0$.


What if we only have the information that $r$ and $s$ are real? Well, if $r=s$ we have $c_2=-b_2r$ and the discriminant of the equation is $0$. Thus $$ (b_1+ib_2)^2-4(c_1+ic_2)=0 $$ or $$ b_1^2-b_2^2-4c_1=0 \qquad\text{and}\qquad 2b_1b_2-4c_2=0. $$ Since $c_2=-b_2r$, we get from the imaginary part $$ b_1b_2+2b_2r=0 $$ that means $b_2=0$ (and $c_2=0$) or $$ b_1=-2r. $$ Then, from $r^2+b_1r+c_1=0$ we get $r^2-2r^2+c_1=0$, or $c_1=r^2$. Plugging in the discriminant, we have $$ 4r^2-b_2^2-4r^2=0 $$ or $$ b_2^2=0 $$ and, again $b_2=0$, from which $c_2=0$.


Much easier solution

The only monic polynomial having $r$ and $s$ as its roots is $$ z^2-(r+s)z+rs $$ so $b=-(r+s)$ and $c=rs$ are real.

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