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If $|z|=2$ and $z_{1}$ and $z_{2}$ be two points on that circle.

Then find maximum value of $\arg(z_{1})-\arg(z_{2})$.

where $z,z_{1},z_{2}$ Represents complex number.

$\theta=\arg(z)$ represents principle argument which means $\theta\in (-\pi,\pi]$.

$(a)\; \pi/2\;\;\; (b)\; \pi\;\;(c)\; 2\pi/3\;\;(d)\; 2\pi$

Try: let $z_{1}=x_{1}+iy_{2}$ and $z_{2}=x_{2}+iy_{2}$ be any two complex number.Then $x^2_{1}+y^2_{1}=4,x^2_{2}+y^2_{2}=4$.

So $\displaystyle \arg(z_{1})-\arg(z_{2})=\arctan(\frac{y_{1}}{x_{1}})-\arctan(\frac{y_{2}}{x_{2}})$.

Could some help me to solve it, Thanks

Community
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DXT
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    What values is arg(z) allowed to take? – Billy Jan 21 '18 at 18:34
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    I presume you are talking about the principal argument. Am I right? – Math Lover Jan 21 '18 at 18:47
  • What is your definition of arg? For $z=a+bi$, it is not usually the case that $\arg (z)$ is defined to be $\arctan\left(\frac{b}{a}\right)$. If this were the definition, $\arg(1+i)$ and $\arg(-1-i)$ would be equal. – Steve Kass Jan 27 '18 at 02:21
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    $\theta =\arg(z)$ represent principle argument. Means $\theta\in (-\pi,\pi]$ – DXT Jan 27 '18 at 03:16

2 Answers2

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The options $(a)(b)(c)$ can be excluded since for $z_1=-\sqrt 3+i$ and $z_2=-\sqrt 3-i$, we have $$\arg(z_1)-\arg(z_2)=\frac{5}{6}\pi-\left(-\frac 56\pi\right)=\frac 53\pi$$ which is larger than $\pi$.

We have $$-\pi\lt\arg(z_1)\le\pi$$ and $$-\pi\lt\arg(z_2)\le\pi\iff -\pi\le -\arg(z_2)\lt \pi$$ from which $$-2\pi\lt \arg(z_1)-\arg(z_2)\lt 2\pi$$ follows.

So, the option $(d)$ is not correct.

Hence, there are no correct options.

($\arg(z_1)-\arg(z_2)$ has no maximum value.)

mathlove
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  • Sir maximum value will be strictly less than 2pi , but there exists a maximum which we are just not sure what unqiue value it is right ? Its very close to 2pi but not exact 2pi ? – Orion_Pax May 17 '22 at 10:54
  • @Orion_Pax : No, it is not right. Suppose that $\arg(z_1)-\arg(z_2)$ has maximum value $M$. Then, taking $\arg(z_1)=\pi$ and $\arg(z_2)=-\frac M2$, we see that $\arg(z_1)-\arg(z_2)=\pi+\frac M2$ is larger than $M$. This is a contradiction. So, $\arg(z_1)-\arg(z_2)$ has no maximum value. – mathlove May 17 '22 at 15:03
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use that $$\arctan(x)-\arctan(y)=\arctan\left(\frac{x-y}{1+xy}\right)$$ for $$xy>-1$$

Dr. Sonnhard Graubner
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