Find limlimits_(z→ i)(z^2+1)/(z^6+1)

Find $\ \lim\limits_{z\to i}\dfrac{z^2+1}{z^6+1}$.

Step 1 · Confirm the indeterminate form

At $z=i$: $\ i^2+1=0$, and $\ i^6+1=(i^2)^3+1=-1+1=0$.

Step 2 · Factor via sum of cubes

Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=z^2,\ b=1$:

$$ z^6+1=(z^2)^3+1^3=(z^2+1)(z^4-z^2+1) $$

Step 3 · Cancel

$$ \frac{z^2+1}{z^6+1}=\frac{z^2+1}{(z^2+1)(z^4-z^2+1)}=\frac1{z^4-z^2+1} \qquad(z\neq\pm i) $$

Step 4 · Evaluate at $z=i$

Using $i^2=-1,\ i^4=1$:

$$ \boxed{\lim_{z\to i}\frac1{z^4-z^2+1}=\frac1{1-(-1)+1}=\frac13} $$

Verification

Expanding back: $(z^2+1)(z^4-z^2+1)=z^6-z^4+z^2+z^4-z^2+1=z^6+1$ ✓. Also $z=i$ is a root of $z^2+1$ but not of the remaining factor: $i^4-i^2+1=1+1+1=3\neq0$, confirming the cancellation was legitimate (the singularity is removable, not a genuine pole).