We need to evaluate the limit:

$ L = \lim_{z \to e^{\frac{\pi i}{3}}} \frac{z^3 + 1}{z^4 + z^2 + 1} $

First, let's substitute $z = e^{\frac{\pi i}{3}}$ into the expression.

Numerator: $z^3 = (e^{\frac{\pi i}{3}})^3 = e^{\pi i} = \cos(\pi) + i \sin(\pi) = -1$. So, $z^3 + 1 = -1 + 1 = 0$.

Denominator: $z^2 = (e^{\frac{\pi i}{3}})^2 = e^{\frac{2\pi i}{3}} = \cos(\frac{2\pi}{3}) + i \sin(\frac{2\pi}{3}) = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$.

$z^4 = z \cdot z^3 = e^{\frac{\pi i}{3}} \cdot (-1) = -(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$.

So, $z^4 + z^2 + 1 = (-\frac{1}{2} - i \frac{\sqrt{3}}{2}) + (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) + 1 = -1 + 1 = 0$.

Since we have the indeterminate form $\frac{0}{0}$, we can use factorization or L'Hôpital's Rule.

Limit Factorization Strategy

Let's factorize the numerator and the denominator.

• Numerator: $z^3 + 1 = (z+1)(z^2 - z + 1)$.
• Denominator: $z^4 + z^2 + 1 = (z^2+1)^2 - z^2 = (z^2+1-z)(z^2+1+z) = (z^2 - z + 1)(z^2 + z + 1)$.

The expression becomes:

$ \frac{(z+1)(z^2 - z + 1)}{(z^2 - z + 1)(z^2 + z + 1)} $

The roots of $z^2 - z + 1 = 0$ are $z = \frac{1 \pm \sqrt{1-4}}{2} = \frac{1 \pm i\sqrt{3}}{2}$, which correspond to $e^{\pm i\pi/3}$. Since the limit is taken as $z \to e^{\frac{\pi i}{3}}$, $z$ is approaching a root of $z^2 - z + 1$. However, for $z \neq e^{\frac{\pi i}{3}}$ in the neighborhood of the limit point, $z^2 - z + 1 \neq 0$. Thus, we can cancel the common factor $(z^2 - z + 1)$:

$ L = \lim_{z \to e^{\frac{\pi i}{3}}} \frac{z+1}{z^2 + z + 1} $

Limit Evaluation with Complex Numbers

Now, substitute $z = e^{\frac{\pi i}{3}}$ into the simplified expression.

• $z = e^{\frac{\pi i}{3}} = \cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3}) = \frac{1}{2} + i \frac{\sqrt{3}}{2}$.
• Numerator: $z+1 = (\frac{1}{2} + i \frac{\sqrt{3}}{2}) + 1 = \frac{3}{2} + i \frac{\sqrt{3}}{2}$.
• $z^2 = (e^{\frac{\pi i}{3}})^2 = e^{\frac{2\pi i}{3}} = \cos(\frac{2\pi}{3}) + i \sin(\frac{2\pi}{3}) = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$.
• Denominator: $z^2 + z + 1 = (-\frac{1}{2} + i \frac{\sqrt{3}}{2}) + (\frac{1}{2} + i \frac{\sqrt{3}}{2}) + 1 = 1 + i\sqrt{3}$.

The limit becomes:

$ L = \frac{\frac{3}{2} + i \frac{\sqrt{3}}{2}}{1 + i\sqrt{3}} $

Multiply the numerator and denominator by 2:

$ L = \frac{3 + i \sqrt{3}}{2(1 + i\sqrt{3})} = \frac{3 + i \sqrt{3}}{2 + i 2\sqrt{3}} $

To simplify, multiply by the conjugate of the denominator $(2 - i 2\sqrt{3})$:

$ L = \frac{(3 + i \sqrt{3})(2 - i 2\sqrt{3})}{(2 + i 2\sqrt{3})(2 - i 2\sqrt{3})} $

Numerator calculation:

$ (3)(2) + (3)(-i 2\sqrt{3}) + (i \sqrt{3})(2) + (i \sqrt{3})(-i 2\sqrt{3}) $ $ = 6 - i 6\sqrt{3} + i 2\sqrt{3} - i^2 (2 \cdot 3) $ $ = 6 - i 4\sqrt{3} - (-1)(6) = 6 - i 4\sqrt{3} + 6 = 12 - i 4\sqrt{3} $

Denominator calculation:

$ (2)^2 - (i 2\sqrt{3})^2 = 4 - (i^2 \cdot 4 \cdot 3) = 4 - (-1 \cdot 12) = 4 + 12 = 16 $

So the limit is:

$ L = \frac{12 - i 4\sqrt{3}}{16} $

Simplify by dividing the numerator and denominator by 4:

$ L = \frac{3 - i \sqrt{3}}{4} = \frac{3}{4} - i \frac{\sqrt{3}}{4} $