I'm trying to find the image of the domain $D$ under the mapping $w=\frac{2z+1}{z+i}$, where
$$D=\left\{z;|z|<1,0<\operatorname{Arg}z<\frac\pi4\right\}$$
I've tried two ways but in both I arrived to something very complex.
First way: Since $z=\frac{-iw+1}{w-2}$, we have $\left|\frac{-iw+1}{w-2}\right|<1$ and after some manipulation we arrive to $4u+2v<3$. Also we should have
$$0<\arctan(\frac{-u^2-v^2+2u-v}{u-2v-2})<\frac\pi4$$
Now if $u-2v-2>0$ we have $$0<\frac{-u^2-v^2+2u-v}{u-2v-2}<1$$ and if $u-2v-2<0$ we should have $-u^2-v^2+2u-v<0$
Second way: Let $w_1=z+i$. Then
$w_1=\left\{z;|z-i|<1,0<\text{Arg}(z-i)<\frac\pi4\right\}$. If $w_2=\frac1{w_1}$, then since $|z-i|<1$ we arrive to $2v+1<0$ and since $0<\text{Arg}(z-i)<\frac\pi4$ we arrive to $0<\frac{-v-u^2-v^2}u<1$ so again we should consider two cases $u>0$ and $u<0$.
Could anyone help me to find the right image?