91Ó°ÊÓ

Expression from Cauchy Riemann theorem:

w = u + iv And z = x + iy, w = f(z) and z(x,y)n, u(x,y) and u(x,y) .

Implicit formula:

$\frac{dw}{dt}=\frac{dw}{dz}.\frac{dz}{dt}=f\left(z\right)\frac{dz}{dt}$

Consider the transformation w = f(z) , where f(z) is analytic at z₀ and $f\left(z_0\right)≠0$.

To prove that under this transformation the tangent at z₀ is rotated through the angle argf(z₀) .

As a point moves from z₀ to $z_0+∆z$ along C the image point moves along in the w-plane from w₀ to $w_0+∆w$ .

If the parameter used to describe the curve is t, then the corresponding to the path z = z(t) in the z-plane, the path w = w(t) in the w-plane. the derivatives$\frac{dz}{dt}·\frac{dw}{dt}$represent the tangent vectors to corresponding on C.

Now$\frac{dw}{dt}=\frac{dw}{dz}·\frac{dz}{dt}=f\left(z\right)\frac{dz}{dt}$and in particular at z₀ and w₀.

${\overline{)\frac{dw}{dt}}}_{w=w_0}=f\text{'}\left(z_0\right){\overline{)\frac{dz}{dt}}}_{z=z_0}$ ...... (1)

Providing f(x) is analytic at z = z₀.

Obtain:

$$\begin{gathered} {\overline{)\frac{dw}{dt}}}_{w=w_0}={\mathrm{ÒÏ}}_0{e}^{i\mathrm{Ï•}}f\text{'}\left(z\right)=R{e}^{i∞} \\ {\overline{)\frac{dz}{dt}}}_{z=z_0}=r_0{e}^{i{\mathrm{θ}}_0}{\mathrm{ÒÏ}}_0{e}^{i\mathrm{Ï•}} \\ {\overline{)\frac{dz}{dt}}}_{z=z_0}=R{r}_0{e}^{i\left({\mathrm{θ}}_0+∞\right)} \end{gathered}$$ .....(2)

So that as required:

${\mathrm{Ï•}}_0={\mathrm{θ}}_0+\mathrm{Î\pm }={\mathrm{θ}}_0+argf\text{'}\left(z_0\right)$ ...... (3)

Note that f(z₀) = 0 then $\mathrm{Î\pm }$ is indeterminate.

Points where f(z) = 0 are called critical points.