E sad, zanima me je li dobro odradjeno, narocito dio nakon eksponencijalnog izraza jer nisam siguran da se tako preslikava.
[dispmath]f(z)=\frac{2i+ie^{2iz}-1}{e^{2iz}+i}\\
D=\left\{z\in\mathbb{C}\mid\text{Im }z<0;\;-\frac{\pi}{2}<\text{Re }z<-\frac{\pi}{4}\right\}[/dispmath]
[dispmath]f(z)=\frac{2i+i\left(e^{2iz}+i\right)}{e^{2iz}+i}=\frac{2i}{e^{2iz}+i}+i\\
z_1=2iz\\
z_2=e^{z_1}\\
z_3=z_2+i\\
z_4=\frac{1}{z_3}\\
z_5=2iz_4\\
\underline{z_6=z_5+i}[/dispmath][dispmath]\begin{matrix}
1^\circ\\
\\
\\
\\
\end{matrix}\qquad\begin{matrix}
\displaystyle z_1=2iz\\
\displaystyle x_1+iy_1=2ix-2y\\
\displaystyle\underline{x_1=-2y}\quad\underline{y_1=2x}\\
\\
\end{matrix}\qquad\begin{matrix}
\displaystyle\text{Im }z<0\\
\displaystyle y=-\frac{x_1}{2}\\
\displaystyle-\frac{x_1}{2}<0\;\Longrightarrow\;\underline{x_1>0}
\end{matrix}\qquad\begin{matrix}
\displaystyle-\frac{\pi}{2}<\frac{y_1}{2}<-\frac{\pi}{4}\\
\displaystyle\underline{-\pi<y_1<-\frac{\pi}{2}}\\
\\
\end{matrix}[/dispmath]
[dispmath]2^\circ\quad z_2=e^{z_1}\\
x_2+iy_2=e^{x_1+iy_1}\\
x_2+iy_2=e^{x_1}(\cos y_1+i\sin y_1)\\
x_2=e^{x_1}\cos y_1\\
\underline{y_2=e^{x_1}\sin y_1}\\
x_2^2+y_2^2=e^{2x_1}\\
|z_2|=\sqrt{e^{2x_1}}=e^{x_1}\\
x_1>0\\
x_1:\;0\to+\infty\\
e^0=1\\
\underline{e^{+\infty}=+\infty}[/dispmath][dispmath]D_2=\mathbb{C}\setminus\{|z_2|<1\}[/dispmath]
[dispmath]\begin{matrix}
3^\circ\\
\\
\\
\\
\\
\end{matrix}\qquad\begin{matrix}
\displaystyle z_3=z_2+i\\
\displaystyle x_3+iy_3=x_2+iy_2+i\\
\displaystyle x_3+iy_3=x_2+i(y_2+1)\\
\displaystyle\underline{x_3=x_2}\\
\displaystyle\underline{y_3=y_2+1}
\end{matrix}\qquad\begin{matrix}
\displaystyle z_2=z_3-i\\
\displaystyle\underline{|z_3-i|<1}\\
\\
\\
\\
\end{matrix}[/dispmath][dispmath]D_3=\mathbb{C}\setminus\{|z_3-i|<1\}[/dispmath]
[dispmath]\begin{matrix}
4^\circ\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\\
\end{matrix}\qquad\begin{matrix}
\displaystyle z_4=\frac{1}{z_3}\\
\displaystyle x_4+iy_4=\frac{1}{x_3+iy_3}\\
\displaystyle x_4+iy_4=\frac{x_3}{x_3^2+y_3^2}+i\frac{-y_3}{x_3^2+y_3^2}\\
\displaystyle x_4=\frac{x_3}{x_3^2+y_3^2};\quad y_4=\frac{-y_3}{x_3^2+y_3^2}\\
\displaystyle x_4^2+y_4^2=\frac{1}{x_3^2+y_3^2}\\
\displaystyle x_3=\frac{x_4}{x_4^2+y_4^2};\quad y_3=\frac{-y_4}{x_4^2+y_4^2}
\end{matrix}\qquad\begin{matrix}
\displaystyle|x_3+i(y_3-1)|<1\\
\displaystyle\sqrt{x_3^2+(y_3-1)^2}<1\\
\displaystyle x_3^2+(y_3-1)^2<1\\
\displaystyle\frac{x_4^2}{\left(x_4^2+y_4^2\right)^2}+\left(\frac{-y_4}{x_4^2+y_4^2}-1\right)^2<1\\
\displaystyle\frac{x_4^2}{\left(x_4^2+y_4^2\right)^2}+\frac{y_4^2}{\left(x_4^2+y_4^2\right)^2}+\frac{2y_4}{x_4^2+y_4^2}+\cancel1<\cancel1\\
\displaystyle\left(x_4^2+y_4^2\right)+2y_4\left(x_4^2+y_4^2\right)<0\quad\Bigr/:\left(x_4^2+y_4^2\right)\\
\\
\end{matrix}[/dispmath][dispmath]1+2y_4<0\\
2y_4<-1\\
\underline{y_4<-\frac{1}{2}}[/dispmath][dispmath]D_4=\left\{z_4\in\mathbb{C}\mid y_4>-\frac{1}{2}\right\}[/dispmath]
[dispmath]5^\circ\quad z_5=2iz_4\\
x_5+iy_5=2ix_4-2y_4\\
x_5=-2y_4;\quad y_5=2x_4\\
y_4=-\frac{x_5}{2}>-\frac{1}{2}\\
\underline{x_5<1}[/dispmath]
[dispmath]6^\circ\quad z_6=z_5+i\\
x_6+iy_6=x_5+iy_5+i\\
x_6+iy_6=x_5+i(y_5+1)\\
x_6=x_5;\quad y_6=y_5+1\\
x_5<1\\
x_6<1[/dispmath][dispmath]f(D)=\{z_6\in\mathbb{C}\mid\text{Re }z_6<1\}[/dispmath]