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We all know that -arg(z)=$arg(\overline{z})$ But what is the relation between arg(-z) and arg(z) ?

Approach if z lies in I and II Quadrant then the relationship that suffice is arg(-z)= -π+arg(z)

Contrary if z lies in III and IV Quadrant then the relationship that suffice is arg(-z)= π+arg(z)

How do we club these two equation so that we get relation between arg(-z) & arg(z)

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I appreciate that if one does a formal solution (such as the arctangent) for $\arg(z)$ and $\arg(-z)$ you'll get the results that you presented. However, when you get right down to it, there is no difference between $\pm\pi$. I mean exactly this:

$$z=|z|e^{i(\theta+\pi)}=|z|e^{i(\theta-\pi)}$$

Therefore you can say that

$$\arg(-z)-\arg(z)=\pm\pi$$

and use either $+\pi$ or $-\pi$. If I had to make a choice I would choose $+\pi$ because that would mean that I was always rotating $z$ anticlockwise to create $-z$.

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