Proving limit of a complex function

Question

how to prove this limit:

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

using epsilon-delta definition of limit?

I tried to make |x+i(2x+y) - (1+i)| in form of |z-(1-i)|, but no luck..

Answer 1

Note that\begin{align}\bigl\lvert x+i(2x+y)-(1+i)\bigr\rvert&=\lvert(x-1)+(2x+y-1)i\rvert\\&=\bigl\lvert(x-1)+\bigl(2(x-1)+(y+1)\bigr)i\bigr\rvert\\&=\sqrt{(x-1)^2+\bigl(2(x-1)+(y+1)\bigr)^2}\end{align}Now, take $\varepsilon>0$. You want to find a $\delta>0$ such that$$\sqrt{(x-1)^2+(y+1)^2}<\delta\implies\sqrt{(x-1)^2+\bigl(2(x-1)+(y+1)\bigr)^2}<\varepsilon.\tag1$$If $\sqrt{(x-1)^2+(y+1)^2}<\delta$, then, since$$\lvert x-1\rvert,\lvert y+1\rvert\leqslant\sqrt{(x-1)^2+(y+1)^2},$$both numbers $\lvert x-1\rvert$ and $\lvert y+1\rvert$ are smaller than $\delta$. And then\begin{align}(x-1)^2+\bigl(2(x-1)+(y+1)\bigr)^2&<\delta^2+9\delta^2\\&=10\delta^2.\end{align}So, take $\delta=\frac\varepsilon{\sqrt{10}}$ and then $(1)$ will hold.