$ f_a = \frac{az + b}{cz +d} $
Would the kernel of this transformation be the scalars of the identity matrix because $ a = d$ & $b=c=0$
so $ \frac{az}d = z$ this leaves us with z or is this wrong.
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Your question is not very clear. What do you mean by the kernel of a Möbius transformation? A Möbius transformation has at most one zero, which will be located at $z = -\frac{b}{a}$, assuming $a \neq 0$. – Thusle Gadelankz Nov 26 '20 at 15:34
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I have something in my groups and symmetries say to take this as a homomorphism from GL(2,C) to Aut(C). It asks me to find the kernel of it. – Balkaran Mali Nov 26 '20 at 15:41
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2Okay. You should include in your question that what you want to find is the kernel of the homomorphism from $\text{GL}(2, \mathbb{C})$ to $\text{Aut}(\mathbb{C}^{*})$, which will be the set of all scalar multiples of the identity matrix, yes! – Thusle Gadelankz Nov 26 '20 at 15:52