Operational definition of conformal
Theorem
If \(f\) is analytic on the region \(A\) and \(f′(z_{0})≠0\), then \(f\) is conformal at \(z_{0}\). Furthermore, the map \(f\) multiplies tangent vectors at \(z_{0}\) by \(f′(z_{0})\).
Proof
The proof is a quick computation. Suppose \(z=γ(t)\) is a curve through \(z_{0}\) with \(γ(t_{0})=z_{0}\). The curve \(γ(t)\) is transformed by \(f\) to the curve \(w=f(γ(t))\). By the chain rule we have
\(\hspace{5cm}\left. \frac{df\left( \gamma \left( t\right) \right) }{dt}\right\vert_{t_{0}}=f^{\prime }\left( \gamma \left( t_{0}\right) \right) \gamma^{\prime }\left( t_{0}\right) =f^{\prime }\left( z_{0}\right) \gamma^{\prime }\left( t_{0}\right) .\)
Example : Basic example
Suppose \(c=ae^{iφ}\) and consider the map \(f(z)=cz\). Geometrically, this map rotates every point by \(φ\) and scales it by \(a\). Therefore, it must have the same effect on all tangent vectors to curves. Indeed, \(f\) is analytic and \(f′(z)=c\) is constant.
Example :
Let \(f(z)=z^{2}\). We have \(f′(z)=2z\). Thus the map \(f\) has a different affect on tangent vectors at different points \(z_{1}\) and \(z_{2}\).
Example : Linear approximation
Suppose \(f(z)\) is analytic at \(z=0\). The linear approximation (first two terms of the Taylor series) is
\(\hspace{5cm}f(z)≈f(0)+f′(0)z\).
If \(γ(t)\) is a curve with \(γ(t_{0})=0\) then, near \(t_{0}\)
\(\hspace{5cm}f(γ(t))≈f(0)+f′(0)γ(t)\).
That is, near \(0\), \(f\) looks like our basic example plus a shift by \(f(0)\).
Example :
The map \(f(z)=\overline{z}\) has lots of nice geometric properties, but it is not conformal. It preserves the length of tangent vectors and the angle between tangent vectors. The reason that it isn't conformal is that it does not rotate tangent vectors. Instead, it reflects them across the \(x-\)axis.
In other words, it reserves the orientation of a pair of vectors. Our definition of conformal map requires that it preserves orientation.