Proof of $\displaystyle \lim_{z\to 1-i}[x+i(2x+y)]=1+i$ I am having some difficulty with the epsilon-delta proof of the limit above.
I know that $|x+i(2x+y)-(1+i)|<\epsilon$ when $|x+iy-(1-i)|<\delta$.
I tried splitting up the expressions above in this way:
$|x+i(2x+y)-(1+i)|\\
=|(x-1)+i(y+1)+i(2x+2)|\\
\le|(x-1)+i(y+1)|+|i(2x+2)|\\
<\delta+|i(2x+2)|$
Is this the correct approach? I don't know how else I can manipulate the expression bounded by $\delta$. Also, what can I do about the i's? 
* $z=x+iy$

Hint: If $z\to 1-i$ and $z=x+iy$, then $x\to 1$ and $y\to -1$.
