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?

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 $z^2=0$
