Question of argument of z and (-z)

Question

I know the relation of $arg (z)$ and $arg(-z)$, but I don't understand the mistake of this reasoning:

If $z=x+iy$, then $arg(z)=\tan^{-1}(y/x)$ so, $-z=-x-iy$, and: $$arg(-z)= \tan^{-1}(-y/-x)=\tan^{-1}(y/x)=arg(z)$$ that has no sence. Please help me with this.

Answer 1

It is true that $$\tan(\arg(z)) = \frac{y}{x}$$

But you cannot deduce in general that $$\arg(z) = \arctan \left( \frac{y}{x} \right)$$

Answer 2

You just observed that the identity

$$\arg(z)=\arctan\left(\frac yx\right)$$

is wrong.

---

$$\arg(z)=\arctan\left(\frac yx\right)+k\pi$$ holds.