Question in Complex Analysis Let $f$ be a analytic in Domain $D$ which is an open disk of radius 1 centered at $z=0$ , it is given that in Domain $D$ , $|f(z)|\leq 1-|z|$ then prove that $f(z)=0$ in Domain $D$

It is asked that we have to prove $f(z) = 0$ using Cauchy's Integral Formula 

I have attached my solution here My solution Please verify this!!
Your solution is correct. A few notes: the second page could be substituted with a reductio ad absurdum: if $f(0)\neq 0$ and $f$ is constant, there is a $z:1-|z|&lt;|f(0)|=|f(z)\le 1-|z|$. 

In general, there's a faster approach to this kind of problems, by means of the cauchy inequality (which is an easy corollary of Cauchy integral formula) : given a function $f$ which is holomorphic in $B(z_0,R)$, we have that, $\forall r&lt;R$:

$$|f^{(n)}(z_0)|\le n! r^{-n}\underset{|z-z_0|=r}{\sup}|f|$$

By Cauchy inequality and the hypotesis, 

$|f^{(n)}(0)|\le n!r^{-n}\underset{|z|=r}{\sup}|f|\le n!r^{-n}(1-r)$

Letting $r\to 1^-$, we obtain

$$f(0)=0\\
\forall_{n&gt;0}f^{(n)}(0)=0$$

Thus we have proved that $f$ is zero on the open unit disc
Let $z\in D$ and $d$ be the distance from $C_r$ to $z$.  

By CIF, $|f(z)|=|1/(2\pi i)\oint_{C_r}f(\zeta)/(\zeta-z)\DeclareMathOperator{\d}{d}\d\zeta|\le\oint_{C_r}|f(\zeta)/(\zeta-z)|\d\zeta\le\oint_{C_r}(1-|\zeta|)/|\zeta-z|\d\zeta\le(1-r)/d\to0$ as $r\to1$.

This is essentially what you did.