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Prove the following proposition

If there exists a branch $F(z)$ of log$z$ in the domain $D$, then any other

branch is of the form $F(z) + 2k$$\pi i$, for some $k \in $ $\mathbf Z$

Conversely $F(z) + 2k$$\pi i$ is a branch of log$z$ for any $k \in $ $\mathbf Z$

Proof

Suppose that $F(z)$ and $G(z)$ are two branch of log$z$, then the difference

$$H(z)=\frac{F(z)-G(z)}{2\pi i}$$

is a continuous function in $D$ which takes only integer values.

Since $D$ is a domain, in particular it is connected, such a function is necessarily constant and we will prove this fact.

The set of points $z\in D$ such that $H(z)$ is equal to a given integer $n$ is both open and closed, and so the set is either empty or equal $D$.

the constant must therefore be an integer.

Thus $F(z) + 2k$$\pi i$ is a branch of log$z$ for any integer $k$ is obvious.

Intuitively i understood what the theorem meant because the branch $H(z)$ have to

go in cycles of $2$$\pi$ and thus there will be a repetition in each cycle but

I dont quite understand the prove above especially from the part onwards

'The set of points $z\in D$ such that $H(z)$ is equal to a given integer...'

Could someone explain this proof to me. Thanks

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Since both $F$ and $G$ are representants of $\log z$ it follows that $$F'(z)=G'(z)={1\over z}\qquad(z\in D)\ .$$ So $H'(z)\equiv0$ in $D$, and as $D$ is connected $H$ has to be constant in $D$.

That this constant has to be of the form $2k\pi i$ with $k\in{\mathbb Z}$ does not follow from your argument. It follows from $e^{F(z)}=e^{G(z)}\equiv z$, or $e^{F(z)-G(z)}\equiv1$.

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