a curious limit

[JmsNxn]

The logarithm of \( z=r e^{i\phi} \) is \( \log( r)+i\phi \).
So regardless how you approach 0, i.e. \( r\to 0 \), you will allways have that \( |\log(z)|=\sqrt{\log( r)^2+\phi^2}\to \infty \).
So the answer is no (except \( 1=e^{vi} \)).

that's what I thought