To use Vieta's formula for complex constant solution or not? 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.
 A: 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.
