Find the value for $v(x,y)$ in the function $f(x+iy)=u(x,y)+iv(x,y)$ with $u(x,y)=\log(x^2+y^2)$

Question

Find the value for $v(x,y)$ in the function $f(x+iy)=u(x,y)+iv(x,y)$ with $u(x,y)=\log(x^2+y^2)$

I used the equations of Cauchy-Riemann and become: $v(x,y)=2 \tan^{-1}(\frac{y}{x})+c$

The answer should be: $-2 \tan^{-1} (\frac{x}{y})+c$

What did I do wrong? To explain my solution, I uploaded a scratch.

Answer 1

Note that $\arctan(x)+\arctan\left(\frac1x\right)=\frac\pi2$. Therefore, $2\arctan\left(\frac xy\right)+2\arctan\left(\frac yx\right)$ is constant.

Answer 2

The function $\, f(x + i y) = 2 \log(x + i y) = \log(x^2 + y^2) + 4 \arctan( y / (x + \sqrt{x^2 + y^2}). \,$ The trouble with using $\, v(x,y)=2 \tan^{-1}(\frac{y}{x})+c \,$ is that the constant $\, c \,$ depends on $\, x<0 \,$ or $\, x>0. \,$ The alternate expression only has a problem when $\, x < 0 \,$ and $\, y = 0. \,$