Existence of continuous and onto function Does there exist a continous and onto function from unit disc to whole complex plane?
I think this is false. 
If we consider its inverse function then domain is set of all complex plane and codomain is unit disc. Therefore, entire function is bounded, hence constant.
Is this correct argument??   Please help. 
 A: Open unit disk in any normed linear space is homeomorphic to the whole space. An explicit heomeomorphism if $ x\to \frac x {1-\|x\|}$. Its inverse is $ x\to \frac x {1+\|x\|}$. Proof: let $y$ be arbitrary and let us try to solve the equation $\frac x {1-\|x\|}=y$. Take norm on both sides to get $\frac {\|x\|} {1-\|x\|}=\|y\|$. This gives $\|x\|=\|y\|-\|x\|\|y\|$ so $\|x\|=\frac {\|y\|} {1+\|y\|}$. Hence $x=(1-\|x\|)y=(1-\frac {|y\|} {1+\|y\|})y=\frac 1 {1+\|y\|}y$. Clearly, this $x$ is in the open unit disk. This proves that that map is onto. For a given $y$, $x$ is unique so the map  is also one-to-one. For the complex plane simply take $|z|$ for the norm. 
A: Yes, such a function exists:
e.g. $f(x+iy) = \frac{x+iy}{1-\sqrt{x^2+y^2}}$
Your argument is invalid for two reasons: You probably ment to ask about holomorph functions (otherwise there is no problem with having the entire complex plain as domain and be bounded).
Second: The function might not have an inverse, therefore you can't easily talk about the inverse.
Edit:
Proof that this function is onto:
First we see that $f$ does not change the argument of a given complex number, i.e. $arg(z) =arg(f(z))$.
Therefore if $f$ is an onto function from $(-1,1)$ to $\mathbb{R}$ (seen as subsets of $\mathbb{C}$), then $f$ is onto in $\mathbb{C}$.
$f$ restircted to $(-1,1)$ is $f(x)=\frac{x}{1-|x|}$. Now let $y\in\mathbb{R}$, then 
\begin{align}
f\left(\frac{y}{1+|y|}\right)&=\frac{\frac{y}{1+|y|}}{1-\left|\frac{y}{1+|y|}\right|}\\
&=\frac{\frac{y}{1+|y|}}{\frac{1+|y|-|y|}{1+|y|}}\\
&=y
\end{align}
