There is no straightline in $C\setminus\{3i\}$ which is mappeed onto a straight line in C by f(True/false) 
Let $\displaystyle f \colon \Bbb C\setminus \{3i\} \to \Bbb C$ be defined by $$f(z)=\frac{z-i}{iz+3}.$$ which of the following statement is true ?

1) f map circles in $C\setminus\{3i\}$ onto circles in C
2)There is no straightline in $C\setminus\{3i\}$ which is mappeed onto a straight line in C by f
My attempts :
Möbius transformation maps circles to circles ..as $C\setminus\{3i\}$ is connected,,as continous image of connected  is connected,,so option 1 is 
trues,...i don't know the option 2,,,pliz help me,,

Hint:
$$
\frac{z-i}{iz+3}=-i+\frac{2}{z-3i}
$$
If $z-3i$ is bounded away from $0$, $f$ is bounded.
