I am attempting to prove the following strictly from the definition of limit:

$$\lim\limits_{z \rightarrow 1-i} [x+i(2x+y)] = 1+i$$

In other words, we want to show that $\forall \varepsilon > 0$, $\exists \delta > 0$ such that:

\begin{align} 0 < |(x+iy)-(1-i)| < \delta &\implies |(x+i(2x+y))-(1+i)|<\varepsilon \\\\ &\iff \\\\ 0 < |(x-1)+i(y+1)| < \delta &\implies |(x-1)+i(2x+y-1)|<\varepsilon \end{align}

I observed that we can rewrite the $\varepsilon$-neighborhood as follows:

\begin{align}|(x-1)+i(2x+y-1)| &= |(x-1)+i(2x-2+y+1)| \\ &= |(x-1)+i(2(x-1)+(y+1))| \\ &= |(x-1)(1+2i)+i(y+1)| \\ &< \varepsilon \end{align}

I am not sure if it means anything that the rewritten $\varepsilon$-neighborhood bears some similarity to the $\delta$-neighborhood (that is, the presence of the $1+2i$ term), but in any case, I am unsure of how to proceed.

  • $\begingroup$ $|x-1| \leq |(x-1)+i(y+1)| < \delta $ $\endgroup$ – Nosrati Mar 4 '17 at 10:32


$|x-1| \le |(x-1)+i(y+1)| < \delta $

$|y+1| \le |(x-1)+i(y+1)| < \delta $

then use triangular inequality with $|(x-1)(1+2i)+i(y+1)|$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.