If $f(z)=(1+i)z+1$, what is $f^4$? Let $f(z)=(1+i)z+1$. Then $f(z)=\sqrt 2 e^{i\pi/4}z+1$ and thus $f=t\circ h\circ r$ where $t$ is the translation of vector $1$, $r$ the rotation of center $0$ and angle $\pi/4$ and $h$ the homothetic of parameter $\sqrt 2$. I found a fix point $z=\frac{-1}{2}+\frac{i}{2}$.
1) What if $f\circ f\circ f\circ f\circ f$ ? It looks to be an translation in the direction of $i$ and an homothetic of parameter 4, but is there an easy way to prove it (with out calculation) ?
2) Let $z_0=i$, $z_1=\sqrt 3+2i$ and $z_2=\sqrt 3-2i$, and consider the triangle $T=\Delta (z_0,z_1, z_3)$. Let $g_n=\underbrace{f\circ ...\circ f}_{n\ times}$. What is the smallest $n$ s.t. $$Area(g_n(T))\geq 100 Area(T)  \ \ ?$$
First, $Area(T)= 2\sqrt 3$. I think that $g_n$ is the same triangle with length of side bigger of $n\sqrt 2$, and thus, I would says that $$Area(g_n(T))=\frac{n4\sqrt 2\cdot n\sqrt 2\sqrt 3}{2}=4\sqrt 3n^2.$$
Therefore, $$4\sqrt 3n^2\geq 200\sqrt 3\implies n^2\geq 50\implies n\geq 8.$$
We conclude that the smallest $n$ is $n=8$. Is it correct ? 
 A: Hint: If $f(z)=az+b$, with $a\ne1$, then $f^n(z)=a^n z + b\dfrac{a^n-1}{a-1}$.
A: The fixed point is $z_0=i$ and rewriting $f$ as 
$$
f(z)-i=(1+i)(z-i)=\sqrt2 e^{i\pi/4}(z-i)
$$ 
you can see that $f$ is a rotation of $\pi/4$ and a homothetic transformation of ratio $\sqrt2$, both with center $z_0$.
It is then obvious that:
$$
f^n(z)-i=(1+i)^n(z-i)=2^{n/2}e^{in\pi/4}(z-i),
$$
that is $f^n$ is a rotation of $n\pi/4$ and a homothetic transformation of ratio $(\sqrt2)^n$, both with center $z_0=i$.
1) For $n=4$, in particular: $f^4(z)-i=-4(z-i)$.
2) As the area scales as the square of a side, you must have $2^{n/2}\ge10$, that is $n\ge7$.
A: 1) The linear part of $f^4$ is $-4z$, therefore, $f^4$ will be of the form $$f^4(z)=-4z+\tau,$$
therefore, it will be an homothetic transformation with scale factor $4$ composed by a point reflexion (you can also say that it's an homothetic transformation with scale factor $-4$) and a translation of vector $\tau$. To find $\tau$, you unfortunately need to do some calculation (i.e. compute explicitly $f^4$).
2) Yes, correct.
