### Instruction: Let $z$ be a complex number such that $|z| = 2.$  Find the largest possible distance between $(3 + 4i)z^3$ and $z^5$ when plotted in the complex plane.### Response: To find the distance between two complex numbers, I need to use the modulus (or absolute value) function.
The modulus of a complex number $a + bi$ is $\sqrt{a^2 + b^2},$ which is the same as the length of the vector from the origin to $a + bi$ in the complex plane.
So, the distance between $(3 + 4i)z^3$ and $z^5$ is $|(3 + 4i)z^3 - z^5|.$
To simplify this expression, I can use the fact that $|z| = 2$ and the properties of the modulus function.
First, I can pull out the factor of $z^3$ from both terms and get $|z^3||3 + 4i - z^2|.$
Then, I can use the fact that $|z^3| = |z|^3 = 2^3 = 8,$ since the modulus function is multiplicative.
So, I have $8|3 + 4i - z^2|.$
Now, I want to maximize this expression, since I'm looking for the largest possible distance.
To do that, I need to maximize the modulus of $3 + 4i - z^2.$
The modulus of a difference of complex numbers is less than or equal to the sum of the moduli of the individual terms, by the triangle inequality.
So, $|3 + 4i - z^2| \leq |3 + 4i| + |z^2|.$
The modulus of $3 + 4i$ is $\sqrt{3^2 + 4^2} = 5,$ by the Pythagorean theorem.
The modulus of $z^2$ is $|z|^2 = 2^2 = 4,$ by the same reasoning as before.
So, $|3 + 4i - z^2| \leq 5 + 4 = 9.$
This means that the largest possible value of $8|3 + 4i - z^2|$ is $8 \cdot 9 = 72.$
Therefore, the largest possible distance between $(3 + 4i)z^3$ and $z^5$ is 72 units.