Find limlimits_(z→1+i)(z^2-z+1-i)/(z^2-2z+2)

Find $\ \lim\limits_{z\to1+i}\dfrac{z^2-z+1-i}{z^2-2z+2}$.

Step 1 · Confirm the indeterminate form

At $z=1+i$: numerator $=(1+i)^2-(1+i)+1-i=2i-1-i+1-i=0$; denominator $=(1+i)^2-2(1+i)+2=2i-2-2i+2=0$.

Step 2 · Factor the denominator

$$ z^2-2z+2=(z-1)^2+1=(z-1)^2-i^2=(z-1-i)(z-1+i) $$

Step 3 · Factor the numerator

Since $z=1+i$ is a root, divide by $(z-1-i)$: matching coefficients of $z^2-z+1-i=(z-1-i)(z+a)$ gives $a=i$, and indeed:

$$ (z-1-i)(z+i)=z^2+iz-(1+i)z-i(1+i)=z^2-z-i+1=z^2-z+1-i \quad\checkmark $$

Step 4 · Cancel and take the limit

$$ \frac{z^2-z+1-i}{z^2-2z+2}=\frac{(z-1-i)(z+i)}{(z-1-i)(z-1+i)}=\frac{z+i}{z-1+i} \qquad(z\neq1+i) $$

$$ \lim_{z\to1+i}\frac{z+i}{z-1+i}=\frac{(1+i)+i}{(1+i)-1+i}=\frac{1+2i}{2i} $$

Step 5 · Simplify

$$ \frac{1+2i}{2i}\cdot\frac{-i}{-i}=\frac{-i-2i^2}{2}=\frac{2-i}2=1-\frac i2 $$

Verification

At $z=0$: original expression gives $\dfrac{1-i}2$; the reduced form gives $\dfrac i{-1+i}=\dfrac{i(-1-i)}{2}=\dfrac{1-i}2$ — matches, confirming the factorization is valid everywhere, not just near $1+i$.