Let z be a complex number such that |z| = 1 — JEE Mathematics

By the triangle inequality:

$$|z^3 - z + 2| \le |z^3| + |-z| + 2$$

Since $|z| = 1$, $|z^3| = |z|^3 = 1$ and $|-z| = |z| = 1$.

$$|z^3 - z + 2| \le 1 + 1 + 2 = 4$$

Let's check if this maximum value can be achieved. For the equality to hold, the complex arguments of $z^3$, $-z$, and $2$ must be equal.
The argument of $2$ is $0$.
For $-z$ to have an argument of $0$, we must have $z = -1$.
Let's test $z = -1$:

$$|(-1)^3 - (-1) + 2| = |-1 + 1 + 2| = |2| = 2 \ne 4$$

Because the phases cannot align perfectly, we must approach it differently.

Let's maximize using algebra:

$$|z^3 - z + 2| = |z(z^2 - 1) + 2| \le |z||z^2 - 1| + 2 = |z^2 - 1| + 2$$

Using the triangle inequality on $|z^2 - 1|$:

$$|z^2 - 1| \le |z^2| + 1 = 1 + 1 = 2$$

So $|z^3 - z + 2| \le 2 + 2 = 4$.

Let's find if a valid $z$ works. Let $z = e^{i\theta}$:
We want to maximize $f(\theta) = |e^{3i\theta} - e^{i\theta} + 2|$.
If we pick $z = i$:

$$|i^3 - i + 2| = |-i - i + 2| = |2 - 2i| = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2} \approx 2.82$$

If we pick $z = -i$:

$$|(-i)^3 - (-i) + 2| = |i + i + 2| = |2 + 2i| = 2\sqrt{2}$$

Let's evaluate more systematically:

$$|z^3 - z + 2|^2 = (z^3 - z + 2)(\bar{z}^3 - \bar{z} + 2)$$

Since $z\bar{z} = 1 \implies \bar{z} = \frac{1}{z}$:

$$= (z^3 - z + 2)\left(\frac{1}{z^3} - \frac{1}{z} + 2\right)$$

Upon optimization, the actual maximum occurs when $Re(z) = -1/2$.
Let $z = -\frac{1}{2} + \frac{\sqrt{3}}{2}i = \omega$.

$$z^3 = 1$$

$$|1 - \omega + 2| = |3 - \omega| = \left|3 - \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right)\right| = \left|\frac{7}{2} - \frac{\sqrt{3}}{2}i\right| = \sqrt{\frac{49}{4} + \frac{3}{4}} = \sqrt{\frac{52}{4}} = \sqrt{13} \approx 3.605$$

It turns out $\sqrt{13}$ is the absolute maximum value.

Answer: $\sqrt{13}$