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 next.

2017年9月4日 · 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)

2014年4月13日 · 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 $\operatorname{Arg}$ or $\operatorname{arg}$

2015年1月23日 · 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 $\mathbb{S^1}$), which corresponds to the fact that the above argument function $\arg$ of any branch of $\log$ (with domain $\mathbb{C} - \{0\}$ anyway) satisfies $$\arg \frac{z}{w} \equiv ...

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. Thanks. EDIT: Just to avoid any confusions I will add how $\arg z$ is defined. $\arg z=\{\text{Arg}\, z+2\pi k\mid k\text{ is an integer}\}$

2018年10月12日 · $\begingroup$ 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 bring up a good point, which is that you may need to add or subtract $2\pi$ to get the argument in the usual range $\endgroup$

2023年7月23日 · If you want a more formal way of doing this, you could rewrite the condition as $$0=\arg(z+i)-\arg(z-1)=\arg\left(\frac{z+i}{z-1}\right)=\arg\left(\frac{(z+i)(z^*-1)}{|z-1|^2}\right).$$ The denominator of this is real and positive, so doesn't affect the argument, so we require $$0=\arg\left((z+i)(z^*-1)\right)=\arg\left((x^2+y^2-x+y)+i(x-1-y ...

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2017年10月1日 · 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 angles that w

2022年8月5日 · Stack Exchange Network. Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.