How will the image of upper half plane $\mathrm{Im}(z)>0$ be under the LFT $w=\frac{3z+i}{-iz+1}$? How can I find 


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*The image of the upper half plane $\mathrm{Im}(z)>0$ under the linear fractional transformation $w=\dfrac{3z + i}{-iz + 1}$.

*The image of the set {${z∈C-0:\{\mathrm{Im}(z)} = \mathrm{Re}(z)\}$} under $w=z + \dfrac{1}{z}$.


For 1.,
I consider $y>0, x \in R$ and then find the inverse $z=\dfrac{w-i}{3+iw}$. This gave me $x=\dfrac{2u}{{(3-v)}^2+u^2}$ and $y=\dfrac{-v^2-u^2+4v-3}{(3-v)^2+u^2}$. Then since $y>0$, 
${-v^2-u^2+4v-3}>0$ which gave me ${(v-2)}^2+u^2<1$.
I therefore knew that the image is the interior of the unit disk centered at $(0,2)$ and drew it. So I need confirmation.
But for the second one, I am confused as how the image should be because if $\text{Im}(z)=\text{Re}(z)$ and $w=z+\dfrac{1}{z}$, then $z$ should not be $0$, but I thought the the line $\text{Im}(z)=\text{Re}(z)$ must pass through the origin. I am waiting for help!
 A: Nice job on the first part. For the second part, note that we're considering the set $$\left\{z\in\Bbb C-0:\text{Re}(z)=\text{Im}(z)\right\},$$ and not the set $$\left\{z\in\Bbb C:\text{Re}(z)=\text{Im}(z)\right\},$$ so we don't have to worry about $z=0$. Now, suppose $z$ is in this set, meaning $z=(1+i)t$ for some non-zero real $t$. Then $$w=z+\frac1z=z+\frac{\overline z}{|z|^2}=(1+i)t+\frac{(1-i)t}{2t^2}=\frac{2t^2+1}{2t}+i\frac{2t^2-1}{2t}.$$ We could express the image as the set of points of the above form, but that isn't very revealing.
Instead, note that if $w=u+iv$ has the above form, then $u+v=2t$ and $u-v=\frac1t,$ so $u^2-v^2=2$. Conversely, if $u^2-v^2=2$ with $u,v$ real then $u\ne v$ and $$\frac{u+v}2=\frac1{u-v}.$$ Putting $t:=u-v,$ then, we have that $t$ is non-zero real, and putting $z=(1+i)t$, we have $w=z+\frac1z,$ as desired. Thus, our image is the hyperbola $$\left\{w:\bigl(\text{Re}(w)\bigr)^2-\bigl(\text{Im}(w)\bigr)^2=2\right\}.$$
A: Just another way for part one (we can use geometry view):
First, note that $$w=B(z)=\dfrac{3z + i}{-iz + 1}=3i+\frac{2}{z+i}$$
So we can write $B$ as $B(z)=B_4\circ B_3 \circ  B_2\circ B_1(z)$ where $B_1(z)=z+i$ translation for $i$, $B_2(z)=\dfrac{1}{z}$ inversion function, $B_3(z)=2z$ homotety with index $2$ and $B_4(z)=z+3i$ translation for $3i$.
Now, first $B_1(\mathbb{H})=\mathbb{H}_1$ where $\mathbb{H}=\{z \in \mathbb{C}: \operatorname{\mathfrak{Im}}(z)>0 \}$ and $\mathbb{H}_1=\{z \in \mathbb{C}: \operatorname{\mathfrak{Im}}(z)>1 \}.$ Then $B_2(\mathbb{H}_1)=\mathbb{D}_1$ where $\mathbb{D}_1 = \{z \in \mathbb{C}: \left|z+\frac{i}{2}\right|<\frac{1}{2}\}$ (we can see this because $B_2(\infty)=0, B_2(i)=-i$ and $B_2(2i)=-\dfrac{i}{2}$ (this third one to see its exterior or interior of disc)), then $B_3(\mathbb{D}_1)=\mathbb{D}_2$ where $\mathbb{D}_2 = \{z \in \mathbb{C}: \left|z+i\right|<1\}$ and then $B_4(\mathbb{D}_2)=\mathbb{D}_3$ where $\mathbb{D}_3 = \{z \in \mathbb{C}: \left|z-2i\right|<1\}.$ 
So $B(\mathbb{H})=\mathbb{D}_3.$
