Number of ordered pairs $(a,b)$ of real numbers satisfying certain conditions with a polynomial How many ordered pairs $(a,b)$ of real numbers are there such that $$(2a^2+1)+(2a^2-1)i$$ is a solution to the equation $$x^2-10x+b^2+4b+20=0$$ There should be $6$.
Here is what I have so far:
$$\begin{align}f(x)&=x^2-10x+b^2+4b+20\\&=(x-5)^2+(b+5)(b-1)\\&=0\end{align}$$
and
$$\begin{align}f(x)&=(x-2a^2-1-(2a^2-1)i)(x-z)\\&= x^2-x\big[2a^2+1+(2a^2-1)i+z\big]+z\big[2a^2+1+(2a^2-1)i\big]\\&=0\end{align}$$
where $z=c+di$ and $z\in\mathbb{C}$.
Comparing coefficients,
$$2a^2+1+(2a^2-1)i+z=10\tag{*}$$
and
$$z\big[2a^2+1+(2a^2-1)i\big]=b^2+4b+20\tag{**}$$
This leads me to three cases: $\mathfrak{Re}(z)=0$ and $\mathfrak{Im}(z)\neq 0$, both $\mathfrak{Re}(z)$ and $\mathfrak{Im}(z)$ are non-zero, or $\mathfrak{Im}(z)=0$ and $\mathfrak{Re}(z)\neq 0$.
Case $1$: $\mathfrak{Re}(z)=0$ and $\mathfrak{Im}(z)\neq 0$
Now $z=c+di=0+di=di$ so by $(*)$, we have $a=\pm\frac{3}{\sqrt{2}}$ and $d=-8$.
Then from $(**)$, we have that $b^2+4b+20=64-80i$ and solving for $b$ yields non real solutions. So this case has no pairs$(a,b)$ of real solutions.
Case $2$: $\mathfrak{Re}(z)$ and $\mathfrak{Im}(z)$ are non-zero. Then, $z=c+di$ and from $(*)$, $c=9-2a^2$ and $d=1-2a^2$. Now from $(**)$, we get that $a=\pm\frac{1}{\sqrt{2}}$ or $a=\pm\sqrt{2}$.
If $a=\pm\sqrt{2}$, then the imaginary part of $(**)$ will not be zero and no real $b$ can be found.
If $a=\pm\frac{1}{\sqrt{2}}$, then this implies $c=8$ and $d=0$ so that $$16=b^2+4b+20\Longrightarrow b=2$$
This case gives the solutions $(\frac{1}{\sqrt{2}}, 2)$ and $(-\frac{1}{\sqrt{2}}, 2)$.
Case $3$: $\mathfrak{Im}(z)=0$ and $\mathfrak{Re}(z)\neq 0$. Now, $z=c\in\mathbb{R}$. This yields the same $2$ solutions as case $2$.
Some how I am missing solutions. Where am I going wrong with this? Am I missing something with symmetry, like if $(2a^2+1)+(2a^2-1)i$ is a solution, then the complex conjugate, $(2a^2+1)-(2a^2-1)i$ must also be a solution and $f(x)$ has a negative discriminant? Also, I was wondering what the fastest method for solving this problem is. Possibly something with group theory? Thanks in advance.

Case 1. $2a^2-1=0$. Then $a=\pm 1/\sqrt{2}$. For each of these $a$, plug  $2a^2+1=2$ into the equation and find  $b=-2$ -we get  $2$ solutions $(a,b)$: $(\frac{1}{\sqrt{2}},-2)$ and $(\frac{-1}{\sqrt{2}}, -2)$.
Case 2. $2a^2-1\ne 0$, then the number $(2a^2+1)+(2a^2-1)i$ is not real. So the number $(2a^2+1)-(2a^2-1)i$ is also a root of the equation. By Vieta, the sum of the roots is 10. So $4a^2+2=10$, $a=\pm \sqrt{2}$. So the root is $5+3i$. Plug  this root  into the equation $(5+3i)^2-10(5+3i)+b^2+4b+20=0.$
Get $b^2+4b-14=0$, get two solutions for $b$.  So we get 4 pairs $(a,b)$ this way, and 6 pairs altogether.
