Show that $f(z)=z^2-iz+2$ is continuous at $z_0=1-i$ by using $(\epsilon,\delta)$
My Attempt: for given $\delta$,if ||(x,y)-(1,-1)||<$\delta$, we have to showed correctly :|x-y|<$\delta$ and |y+1|<$\delta$
Now |f(x,y)-f(1,-1)|=$|x^2-y^2+2ixy-ix-y+2+3i-3|$ i cant process ding further can any one help me
Let us denote $\tilde z = 1-i$ and take $y = \tilde z + \Delta $, a neighbour point of $~\tilde z$. Then, $$ \begin{array}[rcl] ~f(y) - f(\tilde z) &=& f(\tilde z+ \Delta) - f(\tilde z) \\ &=& (\tilde z+\Delta)^2 - i(\tilde z+\Delta) +2 - f(\tilde z) \\ &=& \underbrace{\tilde z^2 - i \tilde z +2}_{f(\tilde z)} +2(\tilde z -i)\Delta + \Delta^2 - f(\tilde z) \\ &=& (2\tilde z -2i + \Delta)\Delta. \end{array} $$ Hence, $$ |f(y) - f(\tilde z)| = |2\tilde z -2i+\Delta| |\Delta| \leq (|2\tilde z -2i|+|\Delta|)|\Delta| , $$ Hence given $\varepsilon>0 $, if you choose $\delta <1 $ and $\delta< \frac{\varepsilon}{|2\tilde z -2i|} $ then you have
$$ |f(\tilde{z}) - f(y)|< \varepsilon ~\, \text{if}~\, |y-\tilde z| = |\Delta| <\delta,$$ which proves the statement.
Hint: start with $|f(z)-f(z_0)|<\epsilon$ $$|z^2-iz+2-(1-i)^2+i(1-i)-2|<\epsilon\\ |z^2-(1-i)^2-i(z-(1-i))|<\epsilon\\ |(x+iy-(1-i))(x+iy+(1-i))-i(x+iy-(1-i))<\epsilon|$$
$\forall\varepsilon>0, \exists\delta>0$ such that if $0<|z-(1-i)|<\delta$ then \begin{eqnarray} |f(z)-f(1-i)| &=& |f(z)-(1-3i)|\\ &=& |z^2-iz+2-\Big((1-i)^2-i(1-i)+2\Big)|\\ &=& |(z^2-(1-i)^2)-iz+i(1-i)|\\ &=& |\Big(z-(1-i)\Big)\Big(z+(1-i)\Big)-i\Big(z-(1-i)\Big)|\\ &=& |\Big(z-(1-i)\Big)\Big((z-(1-i))-i\Big)|\\ &\leq& \delta(\delta+1)\\ &=& \varepsilon \end{eqnarray} because $|z-(1-i)-i|=|(z-(1-i))-i|\leq|z-(1-i)|+|i|\leq\delta+1$
Consider \begin{array} f|f(z)-f(z_0)|&=&|z^2-iz+2-(z_0^2-iz_0+2)|\\ &=&|z^2-z_0^2-i(z-z_0)|\\ &=&|z-z_0|(z+z_0-i)\\ &\leq&|z-z_0|(|z|+3) \end{array} Can you see something closer to the proof?