## Coursera: Analysis of a Complex Kind. 3.5 Analytic functions

### Jul 21, 2016

** Difficult material **

## Slide 2. Analytic functions with zero derivative

f is analytic on domain D in C

if f ‘ (z) = 0 for all z in D, then f is constant in D (same result as real derivative)

Approach to proof

- Show f is constant on any disk Br(a) contained in D
- Use the fact D is connected to show f is constant in D

## Slide 3 More details of proof

<< diagram >>

We are given f ‘ (z) = 0 in the disk so **all the partial derivatives** = 0

If we take any point a in D and move horizontal to psooint b, the value of the point only depends on the variable x. So from real / 1d definition we can show u(a) = u(b) since between them is zero.

Similarly moving from b vertically to c is zero u(b) to u(c) is constant along this vertical line. So u(b) = u(c) and from previous step u(a) = u(c). So the function u is constant in the disk .

In a similar way, the function v is constant in the disk .

Thus f is constant in .

## Slide 4 Second stage of proof

<< diagram >>

Since can get from any point a to b in D along a sequence of overlapping disks, the function is constant in all neighbouring disks and hence along the path

<< continue >>

Jul 30, 2016 at 10:09 am

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