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? 
* $z=x+iy$
 A: Write
$$
x+i(2x+y)-(1+i)=x-1+i(2(x-1))+i(y+1).
$$
Then use the triangular inequality (twice) to get:
$$
|x+i(2x+y)-(1+i)|\leq |x-1|+2|x-1|+|y+1|=3|x-1|+|y+1|\leq 3(|x-1|+|y+1|).
$$
Finally, use
$$
|a|+|b|\leq \sqrt{2}\sqrt{a^2+b^2}
$$
which reduces to the well-known $2|a||b|\leq a^2+b^2$ when raising to the square.
This yields:
$$
|x+i(2x+y)-(1+i)|\leq 3\sqrt{2}\sqrt{(x-1)^2+(y+1)^2}=3\sqrt{2}|z-(1-i)|.
$$
I think you can conclude from this estimate.
Edit: Now I realize there is a faster route by observing that $|x-1|\leq|z-(1-i)|$ and $|y+1|\leq |z-(1-i)|$ (absolute values of real parts and imaginary parts are both smaller than the modulus).
This yields more directly the inestimate $|expression|\leq 4|z-(1-i)|$.
A: First of all, my first-year calculus teacher taught us a trick (we were supposed to keep it secret, please don't rat me out): The $\epsilon$ - $\delta$ dance says you have to pick $\epsilon$ and show there is a corresponding $\delta$. He told us to start from $\delta$ and derive an expression for $\epsilon$ in terms of $\delta$ on scratch paper, and then work backwards in the clean proof.
Scratch paper
Let's start with $\lvert x - i y - (1 - i) \rvert \le \delta$, and see what this implies for $\lvert x + i (2 x + y) \rvert$. By the triangular inequality, we know that $\lvert x - i y - (1 - i) \rvert \le \lvert x - 1 \rvert + \lvert y + 1 \rvert$, if we make the later less than $\delta$, so is the former. A way of getting this is to have $\lvert x - 1 \lvert \le \delta / 2$ and $\lvert y + 1 \lvert \le \delta / 2$.
Now consider:
$$
\begin{align*}
\lvert x + i (2 x + y) - 1 - i \rvert &= \lvert (x - 1) + i(2 x + y - 1) \rvert \\
   &\le \lvert x - 1 \rvert + \lvert 2 x + y - 1 \rvert \\
   &\le \delta / 2 + \lvert (2 x - 2) + (y + 1) \rvert \\
   &\le \delta / 2 + 2 \cdot \delta / 2  + \delta / 2 \\
   &= 2 \delta
\end{align*}
$$
So for $\epsilon$ you can take $\delta = \epsilon / 2$ (or anything less, as convenient).
Clean proof
Left as an exercise (famous words from textbooks that my fingers where itching to write ;-)
A: Hint: If $z\to 1-i$ and $z=x+iy$, then $x\to 1$ and $y\to -1$.
