Limit in complex space How to prove that the limit in the complex space:
$$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=1$$
Please help me
 A: Hint
$$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=\lim\limits_{z \to 0} \frac{|1+z|^2-1}{z(|1+z|+1)}=\lim\limits_{z \to 0} \frac{(1+z)(1+\bar{z})-1}{z(|1+z|+1)}\\=\lim\limits_{z \to 0} \frac{z+\bar{z}+z\bar{z}}{z(|1+z|+1)}\\=\lim\limits_{z \to 0} \frac{z+z\bar{z}}{z(|1+z|+1)}+\lim\limits_{z \to 0} \frac{\bar{z}}{z}\frac{1}{|1+z|+1}$$
Now the fact that $\lim_{\bar{z}}{z}$ does not exist, and all the other limits in the last step exists and are non-zero gives you that the limit exists. 
A: The limit is not well defined. Denote $z=ae^{i\theta}$, then the special limit
$$\lim\limits_{a \to 0} \frac{|1+z|-1}{z}=\cos(\theta)e^{-i\theta}.$$
You need to specify the way the limit should be performed.
A: $$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=\lim\limits_{(x,y) \to (0,0)} \frac{\sqrt{(x+1)^2+y^2}-1}{x+iy}$$ 
if $x=y$
$$\lim\limits_{z \to 0} \frac{|1+z|-1}{z}=\lim\limits_{x\to 0} \frac{\sqrt{(x+1)^2+x^2}-1}{x+ix}\times \frac{\sqrt{(x+1)^2+x^2}+1}{\sqrt{(x+1)^2+x^2}+1}=\lim\limits_{x\to 0} \frac{{(x+1)^2+x^2}-1}{{(x+ix)(\sqrt{(x+1)^2+x^2}+1)}}=\lim\limits_{x\to 0} \frac{x(2x+2)}{x(1+i)\sqrt{(x+1)^2+x^2}+1}=\frac{2}{(1+i)\sqrt{(0+1)^2+0^2}+1}=\frac{1}{(1+i)} $$ 
 similarly for $y=-x$
$$\lim\limits_{(x,y) \to (0,0)} \frac{\sqrt{(x+1)^2+y^2}-1}{x+iy} =\frac{1}{(1-i)}$$ 
hence the limit doesn't exist.
A: In the following, $t$ will be a real number. We have $\lim_{t\to 0}(|1+t|-1)/t = \lim_{t\to 0} (1+t-1)/t  = 1.$ On the other hand $(|1+it|-1)/(it)$ is always purely imaginary. It follows that the full limit does not exist.
