### Instruction: Let $z$ be a complex number satisfying $z^2 = 4z - 19 + 8i$. Given that $|z|$ is an integer, find $z.$.### Response: I notice that the equation has both real and imaginary parts, so I try to separate them and form a system of equations.
Let $z = a + bi$, where $a$ and $b$ are real numbers.
Then, $z^2 = (a + bi)^2 = a^2 - b^2 + 2abi$.
Equating the real parts, I get $a^2 - b^2 = 4a - 19$.
Equating the imaginary parts, I get $2ab = 8$.
I can simplify the second equation by dividing both sides by $2$, giving $ab = 4$.
I also notice that $|z| = \sqrt{a^2 + b^2}$, which is an integer by the given condition.
I try to find some possible values of $a$ and $b$ that satisfy both equations and the condition.
One obvious pair is $a = 4$ and $b = 1$, which gives $|z| = \sqrt{17}$.
However, this is not an integer, so I discard this pair.
Another possible pair is $a = 2$ and $b = 2$, which gives $|z| = \sqrt{8}$.
This is also not an integer, so I discard this pair as well.
I wonder if there are any other pairs that have integer products and sums of squares.
I recall that the only pairs of squares that add up to a multiple of $5$ are $(0, 5)$, $(3, 4)$, and $(5, 0)$.
I check if any of these pairs have products that are multiples of $4$.
I see that only $(3, 4)$ does, since $3 \cdot 4 = 12$.
I try to see if this pair works for the original equation.
If $a = 3$ and $b = 4$, then $z = 3 + 4i$.
Plugging this into the equation, I get $(3 + 4i)^2 = 4(3 + 4i) - 19 + 8i$.
Simplifying, I get $-7 + 24i = -7 + 24i$, which is true.
Therefore, $z = 3 + 4i$ is a solution that satisfies all the conditions.
I wonder if there are any other solutions.
I realize that if $z$ is a solution, then so is its complex conjugate $\overline{z}$.
This is because $\overline{z}^2 = \overline{z^2} = \overline{4z - 19 + 8i} = 4\overline{z} - 19 - 8i$.
So, another solution is $\overline{z} = 3 - 4i$.
I don't think there are any other solutions, because the equation is quadratic and can have at most two roots.
I check my answer by plugging both solutions into the equation and verifying that they work.