we need to find the points where $|f(z)|$ has maximum and minimum value $f(z)=(z+1)^2$ and $R$ be the triangle with vertices $(0,0),(0,1),(2,0)$, we need to find the points  where $|f(z)|$ has maximum and minimum value
so here $z=2$ is the point of maximum and $z=0$ is minimmum (intuitively), as $f$ is analytic so it must attain maxima minima on boundary, 
but my question is how can I show $f$ has no extrimum value on the other points of the boundary of the triangle? Thank you for help

On the lower edge of the triangle, we have $z = x$, and 
$$|f(z)| = (x + 1)^2$$
This is a strictly increasing function of $x$, and so its minimum is at $0$ and maximum at $1$.

On the vertical edge, we have $z = iy$ and
$$|f(z)| = |(iy + 1)^2| = 1 + y^2$$
Again, this is strictly increasing in $y$ (take a derivative), and so maximized at $y = 1$, but the maximum value is only $2$.

The third edge is defined by the line $y = 1 - \frac{x}{2}$, so
\begin{align*}
|f(z)|^2 &= |1 + z|^2 \\
&=\left| 1 + x + i \left(1 - \frac{x}{2}\right)\right|^2 \\
&= (1 + x)^2 + \left(1 - \frac{x}{2}\right)^2 \\
&= 1 + 2x + x^2 + 1 - x + \frac{x^2}{4} \\
&= 2 + x + \frac{x^2}{4}
\end{align*}
This is strictly increasing in $x$ for $0 \le x \le 2$, and the maximum is obtained at the right endpoint, giving a maximum of $5$, as expected.
