Find $a, b \in \mathbb{R}$ if $a+ia$ is a root of the equation $z^2+4z+b=0$. I managed to solve until $a=-2, b=8$ by expanding the conjugate roots and then comparing coefficients.
However, there is another pair of solution where $a=0,b=0$. What is the explanation for this second pair of solution?
 A: Since coefficient are real root will exist in Conjugate pairs
i.e if a(1+i ) is one of root then other is a(1-i ) only.
Now use sum of roots and product of roots 
a=-2 and b=8
The other solution is conditional if b=0 then the roots are real and equal both 0,0 as equation get reduced to $z^2=0$
A: While it is true that if $a(1+i)$ is a root, then $a(1-i)$ is also a root of the equation, care must be taken that $a(1+i)$ has non-zero imaginary part and is non-real. When $a=0$, the root is real and the result can't be applied.
A: 1)
If $w\in \mathbb C$ is a root of a polynomial with real coefficients then $\overline w$ is a root.  But that doesn't mean $\overline w$ is a different and second root.  
If $w = \overline w$ (i.e. if $w \in \mathbb R$) then it is possible but $w$ and $\overline w = w$ are two representations of the same single root. (or it could be that $w$ is a double root.... but it could be a single root).  
In other words the conjugate root theorem is trivial for real numbers.  As every real number is its own conjugate, then all the theorem states for real roots is.... If $w$ is a real root then... $w$ is a real root.
2)
If $u$ and $v$ are are two different roots, or if $u=v$ but $u$ is a double root, to a quadratic polynomial $z^2 + az + b$ then $(z-u)(z-v) = z^2 +az + b$.  
But if $u$ is only one root and we don't know the other root ... we can't do anything.
....
So.....
If $a+ai$ is one root then $a-ai$ is a root.  And IF $a+ai \ne a-ai$ then, yes, we may conclude that $(z-(a+ai))(z-(a-ai)) = z^2 + 4z + b$ and we may conclude that $a=-2$ and $b=8$.
But if $a+ai = a-ai$ and is a single root, then the other root... is something else (although we can conclude the other root is real).
If we do try to solve this, $a+ai = a-ai \implies a=0$ and if the other root is $u$ the we have $(z-0)(z-u)= z^2 -uz + 0*u = z^2 +4z + b$ and thus $u=-4$ and $b = 0$.
.....
Or in other words.  If $w$ is one root we know that $z-w$ divides $P(z)= z^2 + 4z + b$.  ANd if $\overline w$ is another root we know $z-\overline w$ divides $P(z)$.  And so $(z-w)(z-\overline w)$ divides $P(z)$.  But as $P(z)=z^2 +4z + b$ is of degree $2$ and so is $(z-w)(z-\overline w)$ then $(z-w)(z-\overline w) = P(z)$.
But if $w$ is one real root then $\overline w = w$ and all we know is that $z- w = z-\overline w$ divides $P(z)= z^2 + 4z + b$  but that doesn't mean $(z-w)(z-\overline w) = (z-w)^2$ divides.  We only have one root, not two.
So if $a+ai= a-ai$ then $a = 0$ and $a+ai = 0$. and $z-0 = z$ divides $P(z)=z^2+4z + b$ so $z|b$ so $b = 0$.
If we knew the other root $u$ we'd know $z-u$ divides $z^2 + 4z +0$.  But we know that $\frac {z^2 + 4z+0}{z} = z+4$ we know that $z+4$ divides $P(z)$ and $P(z) = z(z+4)$ and the other root is $u = -4$... but that isn't what was asked.
