Sep 4, 2017 · 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)= …

Im trying to find all set of points on the complex plane for which $|z|=\arg z$. I rewrote $|z|= \sqrt{x^2 + y^2}$ and $\arg z$ as $\tan^{-1}(y/x)$.I set them equal But im not sure what to do …

Apr 13, 2014 · Domain of $\operatorname{Arg}(1/z)$: $\operatorname{Re}(z) \neq 0$ 0 Rotational property of multiplying complex numbers together: wanted to know if $\theta$ referred to the …

Jan 23, 2015 · It's anyway more natural (though often not more convenient) to view the argument as a function that takes values modulo $2\pi$ (equivalently, as elements of $\mathbb{S^1}$), …

Let $θ_1 \in \arg(z)$ and $θ_2 \in \arg(w)$. Then, $θ_1+θ_2 \in \arg(z)+\arg(w)$. Also, $θ_1+θ_2 \in \arg(zw)$. Is this sufficient for the proof or correct at all? Hope someone could help me out. …

Oct 12, 2018 · $\begingroup$ Yes, because $\arg(1/-1)=\arg(-1)=\pi=-\pi=-\arg(-1)$. The angles $\pi$ and $-\pi$ are equivalent. The angles $\pi$ and $-\pi$ are equivalent. Although you do …

Jul 23, 2023 · If you want a more formal way of doing this, you could rewrite the condition as $$0=\arg(z+i)-\arg(z-1)=\arg\left(\frac{z+i}{z-1}\right)=\arg\left(\frac{(z+i)(z^*-1)}{|z …

Apr 25, 2022 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …

Oct 1, 2017 · You are thinking of $\arg (z)$ as a real number modulo $2\pi$ or, equivalently, a point on the unit circle. The source you quote is thinking of $\arg (z)$ as the set of all possible …

Dec 13, 2015 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for …