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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.

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  • $\begingroup$ It'd leave you with, $$\lim_{x,y\to 0}\frac{y}{1+\sqrt{x^2+y^2}}$$ $\endgroup$ – Prasun Biswas Apr 19 '15 at 3:51
  • $\begingroup$ Note $|\text {Im}(u)|\le |u|$ for any complex $u.$ $\endgroup$ – zhw. Apr 19 '15 at 3:52
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$$\frac{z}{1+|z|}=\frac{x+iy}{1+\sqrt{x^2+y^2}}=\frac{x}{1+\sqrt{x^2+y^2}}+\left(\frac{y}{1+\sqrt{x^2+y^2}}\right)i$$

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: $\Im(z)=\textrm{Im}(z)$

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From the definition of a limit, $f(z) \to L$ as $z \to z_0$, if for any $\epsilon >0$, there exists a number $\delta>0$ such that $|f-L|<\epsilon$ whenever $0<|z-z_0|<\delta$.

Here, $f=\text{Im}\left(\frac{z}{1+|z|}\right)=\frac{y}{1+\sqrt{x^2+y^2}}$ and $z_0=0$.

Let's show that we can find a $\delta>0$ so that $|f-0|<\epsilon$ whenever $0<|z|=\sqrt{x^2+y^2}<\delta$.

$$\begin{align} |f(z)|&=\left|\frac{y}{1+\sqrt{x^2+y^2}}\right|\\\\ &\le\frac{\sqrt{x^2+y^2}}{1+\sqrt{x^2+y^2}}\\\\ &\le\frac{\sqrt{x^2+y^2}}{1}\\\\ &=\sqrt{x^2+y^2}\\\\ &<\epsilon \end{align}$$

whenever $\sqrt{x^2+y^2}<\epsilon$.

Thus, given any $\epsilon$, choose $\delta \le \epsilon$ to guarantee that $|f(z)|<\epsilon$ whenever $0<|z|<\delta$.

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