Find limlimits_(z→ e^(iπ/3))(z-e^(iπ/3))dfrac zz^3+1

Find $\ \lim\limits_{z\to e^{i\pi/3}}\left(z-e^{i\pi/3}\right)\dfrac z{z^3+1}$.

Step 1 · Confirm $\omega$ is a root of $z^3+1$

$$ \omega^3=\left(e^{i\pi/3}\right)^3=e^{i\pi}=-1 \ \Longrightarrow\ \omega^3+1=0 $$

Step 2 · Find the other two roots and factor

The roots of $z^3=-1$ are $z_k=e^{i\pi(2k+1)/3}$ for $k=0,1,2$: namely $\omega=e^{i\pi/3},\ -1=e^{i\pi},\ \bar\omega=e^{-i\pi/3}$. So:

$$ z^3+1=(z-\omega)(z+1)(z-\bar\omega) $$

Step 3 · Cancel the shared factor

$$ (z-\omega)\frac z{z^3+1}=(z-\omega)\frac z{(z-\omega)(z+1)(z-\bar\omega)}=\frac z{(z+1)(z-\bar\omega)} \qquad(z\neq\omega) $$

Step 4 · Evaluate at $z=\omega$

With $\omega=\frac12+i\frac{\sqrt3}2,\ \bar\omega=\frac12-i\frac{\sqrt3}2$:

$$ \omega+1=\frac32+i\frac{\sqrt3}2, \qquad \omega-\bar\omega=i\sqrt3 $$

$$ (\omega+1)(\omega-\bar\omega)=\left(\frac32+i\frac{\sqrt3}2\right)(i\sqrt3)=\frac{3i\sqrt3}2-\frac32=\frac32(-1+i\sqrt3) $$

Step 5 · Combine and simplify

$$ \frac\omega{(\omega+1)(\omega-\bar\omega)}=\frac{\frac{1+i\sqrt3}2}{\frac32(-1+i\sqrt3)}=\frac{1+i\sqrt3}{3(-1+i\sqrt3)} $$

Multiplying by the conjugate $\dfrac{-1-i\sqrt3}{-1-i\sqrt3}$: denominator $=3[1-(-3)]=12$; numerator $=(1+i\sqrt3)(-1-i\sqrt3)=-1-2i\sqrt3+3=2-2i\sqrt3$:

$$ \boxed{\frac{2-2i\sqrt3}{12}=\frac{1-i\sqrt3}6} $$

Verification

Direct numerical evaluation of $(z-\omega)\dfrac z{z^3+1}$ at points approaching $\omega=e^{i\pi/3}$ from several directions converges to $0.16667\ldots-0.28868\ldots i$, matching $\dfrac{1-i\sqrt3}6\approx0.1667-0.2887i$ to full precision.