Möbius transform answer check for $0$ to $2$,$-2i$ to $0$, $i$ to $\frac32$

Question

Continuation of this question

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Is this the correct answer for the möbius transformation corresponding to:

$0$ to $2$

$-2i$ to $0$

$i$ to $\frac32$

$$\frac{az+b}{cz+d}\cong \frac{az+2azi}{az+ai}=1+\frac{ai}{az+ai}$$