I am trying to check the continuity of this complex function at the origin.
$f(z)=\begin{cases}
\operatorname{Im}( \frac{z}{1+|z|} ) \qquad &\mbox{when } z\neq0,\\ 0
\qquad &\mbox{when }z=0.
\end{cases}$
According to my understanding (correct me if i am wrong), in order for a this function to be continuous at the origin, first, $f(0)$ must exists!(which it does) Then,the limit of $f(z)$ as it tends to 0 must exists too. And both has to be the same.
so,
$\lim_{z\to0} ( \operatorname{Im} ( \frac{z}{1+|z|} ) ) = \lim_{x\to0 \\ y\to0}(\operatorname{Im} ( \frac{x+iy}{1+\sqrt[]{x^2+y^2} } )) $
But now , if we take only the imaginary part of the function,(i.e $y$ from $iy$), wouldn't it leave only leave $\lim_{x\to0 \\ y\to0} y$ ?
Then if y approach 0 first, then the limit is zero. But if x approach 0 first, nothing happens.
Can someone kindly enlighten me where have I got this wrong? I am learning complex analysis by myself and English is not my first language so, I believe I've got some wrong ideas on how to approach this problem.
Best Answer
$$\frac{z}{1+|z|}=\frac{x+iy}{1+\sqrt{x^2+y^2}}=\frac{x}{1+\sqrt{x^2+y^2}}+\left(\frac{y}{1+\sqrt{x^2+y^2}}\right)i$$
Now, it is trivial that,
$$\Im\left(\frac{z}{1+|z|}\right)=\frac{y}{1+\sqrt{x^2+y^2}}$$
I hope you can do the rest.
Note: $\Im(z)=\textrm{Im}(z)$