Complex analysis problem, prove that $f$ is a polynomial of degree $\leq 1$ 
Suppose that $f$ is entire and that for each $z$, either $|f(z)|≤1$ or $|f′(z)|≤1$. Prove that f is a polynomial of degree $\leq 1$.
Hint: Use a line integral to show that $|f(z)|≤A+|z|$ where$ A = \max\{1,|f(0)|\}$.

This is an exercise of complex analysis (author : newman, bak)
Actually, I found similar question but I don't understand answer.
Suppose that $ f $ is entire and that for each $ z $, either $ |f(z)| \leq 1 $ or $ |f^\prime (z) |\leq 1 $. Prove that $ f $ is a linear polynomial.
Is there someone explaining easily?
 A: Inequality: I'll repeat the argument given by muzzlator here, modulo a minor modification which, I think, is necessary. I'll try also to give a few more details. Of course, muzzlator deserves all the credit. I will also show how to conclude that $f$ is a polynomial of degree $\leq 1$ to really add something to his answer. 
If $z\in\mathbb{C}$ and denote $[0,z]$ the line segment with endpoints $0$ and $z$, parametrized by $\{tz\;;\;t\in[0,1]\}$. Then the set
$$
C:=\{t\in[0,1]\;;\;|f(tz)|\leq 1\}
$$
is closed by continuity of $f$. It is also bounded, so $C$ is compact. 
If $C$ is empty, set $t_0:=0$ so $|f(t_0z)|=|f(0)|>1$ hence $|f(t_0z)|=|f(0)|=\max\{|f(0)|,1\}=A$. 
If not, we set $t_0:=\sup C=\max C$. Note that $|f(t_0z)|\leq 1\leq \max\{|f(0)|,1\}=A$. 
If $t_0=1$, then $|f(z)|=|f(t_0z)|\leq 1 \leq \max\{|f(0)|,1\}=A$. And the inequality you seek follows.
So we have to prove the inequality when $0\leq t_0<1$, in which case $|f(t_0z)|\leq A=\max\{|f(0)|,1\}$. By definition of $t_0$, we have $|f(tz)|>1$ for every $t_0<t\leq 1$. By assumption, this forces $|f'(tz)|\leq 1$ on this interval. And by continuity of $f'$, this holds also at $t_0$. Hence
$$
|f'(tz)|\leq 1\qquad\forall t_0\leq t\leq 1.
$$
Now consider the funtion $g(t):=f(tz)$. It is $C^1$ on $\mathbb{R}$ with derivative $g'(t)=f'(tz)z$. Therefore, by the Fundamental Theorem of Calculus
$$
f(z)=g(1)=g(t_0)+\int_{t_0}^1g'(t)dt=f(t_0z)+z\int_{t_0}^1f'(tz)dt
$$
whence
$$
|f(z)|\leq |f(t_0z)|+|z|\int_{t_0}^1|f'(tz)|dt\leq A+|z|\int_{t_0}^11dt=A+|z|(1-t_0)\leq A+|z|.
$$
Line integral: the integral $\int_{t_0}^1g'(t)dt=\int_{t_0}^1f'(tz)zdt$ is the line integral of the analytic function $f'$ over the line segment $[t_0z,z]$. Up to orientation, this does not depend on the parametrization. Here we took the parametrization $t\mapsto tz$ over $[t_0,1]$, oriented from $t_0z$ to $z$. It is denoted
$$
\int_{[t_0z,z]}f'(w)dw=\int_{t_0}^1f'(tz)zdt.
$$
Be careful, the lhs is not the Lebesgue integral of $f'$ over the zero measure set $[t_0z,z]$ in $\mathbb{C}$, otherwise it would be $0$.
Conclusion:  we can deduce that $f$ is a polynomial of degree $\leq 1$ from the estimate $|f(z)|\leq A+|z|$, knowing that $f$ is entire. 
You can either adapt the proof of Liouville's theorem by proving that the coefficients of the Taylor series representation of $f$ at $0$ are zero for $n\geq 2$ from Cauchy's integral formula.
Or you can observe that the inequality entails that the entire function
$$h(z):=\frac{f(z)-f(0)}{z}\quad \forall z\neq 0\qquad h(0):=f'(0)$$ 
is bounded, whence constant by Liouville. 
