Assume $f$ is analytic on $\left|z-1\right|<1$,$f'(z)=1/z$ and $f(1)=0$, Prove that on this region $f(z)=\text{Log}(z)$.

Question

Assume $f$ is analytic on $\left|z-1\right|<1$,$f'(z)=1/z$ and $f(1)=0$, Prove that on this region $f(z)=\text{Log}(z)$.

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Since $f$ is analytic on $\left|z-1\right|<1$, so for every $z=x+iy$ in this region we have that $u_x(x,y)=v_y(x,y)$ and $u_y(x,y)=-v_x(x,y)$, on the other hand $f'(z)=1/z$ implies that $u_x(x,y)=\frac{x}{x^2+y^2}$ and $v_x(x,y)=\frac{-y}{x^2+y^2}$, So $u(x,y)=1/2 \ln(x^2+y^2)+c_1(y)$, Moreover $$\frac{y}{x^{2}+y^{2}}+c_1'\left(y\right)=\frac{y}{x^{2}+y^{2}}$$

From which we conclude that $c_1(y)=c$ $$v\left(x,y\right)=\arctan\left(\frac{y}{x}\right)+c_{2}\left(y\right)$$

$$\frac{1}{\sqrt{x^{2}+y^{2}}}+c'_{2}\left(y\right)=\frac{x}{x^{2}+y^{2}}$$

Is this the correct way to go?

Answer 1

$u_x=\dfrac{x}{x^2+y^2}$ implies $u(x,y)=\dfrac{1}{2}\ln(x^2+y^2)+c(y)$

whereas $u_y=\dfrac{y}{x^2+y^2}$ implies $u(x,y)=\dfrac{1}{2}\ln(x^2+y^2)+l(x)$

which forces $c(y)$ and $l(x)$ to be equal to the same constant function.

Also,

$v_y=\dfrac{x}{x^2+y^2}$ implies $v(x,y)=\arctan\left(\dfrac{y}{x}\right)+g(x)$

whereas $v_x=\dfrac{-y}{x^2+y^2}$ implies $v(x,y)=\arctan\left(\dfrac{y}{x}\right)+h(y)$, which you can verify by partial-differentiating $\arctan\left(\dfrac{y}{x}\right)$ with respect to $x$.

Answer 2

For $z \in D_1(1) = \{z \in \mathbb{C} : |z - 1| < 1\}$, let $g(z) = f(z) - \log(z)$. Since $\log'(z) = 1/z$ and $\log(1) = 0$, we have $g'(z) = 0$ for all $z \in D_1(1)$ and $g(0) = 0$. Since $D_1(1)$ is connected, this implies that $g = 0$ on $D_1(1)$.