Question 8.3: Consider the transformation w = f(z) where f(z) is analytic ...
Consider the transformation w=f(z) where f(z) is analytic at z_{0} and f^{\prime}\left(z_{0}\right) \neq 0. Prove that under this transformation, the tangent at z_{0} to any curve C in the z plane passing through z_{0} [Fig. 8-71] is rotated through the angle \alpha=\arg f^{\prime}\left(z_{0}\right) [Fig. 8-8].
As a point moves from z_{0} to z_{0}+\Delta z along C, the image point moves along C^{\prime} in the w plane from w_{0} to w_{0}+\Delta w. If the parameter used to describe the curve is t, then corresponding to the path z=z(t) [or x=x(t), y=y(t)] in the z plane, we have the path w=w(t) [or u=u(t), v=v(t) ] in the w plane.
The derivatives d z / d t and d w / d t represent tangent vectors to corresponding points on C and C^{\prime}.
Now
\frac{d w}{d t}=\frac{d w}{d z} \cdot \frac{d z}{d t}=f^{\prime}(z) \frac{d z}{d t}
and, in particular at z_{0} and w_{0},
\left.\frac{d w}{d t}\right|_{w=w_{0}}=\left.f^{\prime}\left(z_{0}\right) \frac{d z}{d t}\right|_{z=z_{0}} (1)
provided f(z) is analytic at z=z_{0}. Writing
\left.\frac{d w}{d t}\right|_{w=w_{0}}=\rho_{0} e^{i \phi_{0}}, \quad f^{\prime}(z)=R e^{i \alpha},\left.\quad \frac{d z}{d t}\right|_{z=z_{0}}=r_{0} e^{i \theta_{0}}
we have from (1)
\rho_{0} e^{i \phi_{0}}=R r_{0} e^{i\left(\theta_{0}+\alpha\right)} (2)
so that, as required,
\phi_{0}=\theta_{0}+\alpha=\theta_{0}+\arg f^{\prime}\left(z_{0}\right) (3)
Note that if f^{\prime}\left(z_{0}\right)=0, then \alpha is indeterminate. Points where f^{\prime}(z)=0 are called critical points.