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How are the complex numbers $z_1$ and $z_2$ related if $\arg(z_1) = \arg(z_2)$?

My attempt :
Let $z_1$,$z_2\in \mathbb{C}$ such that $z_1=r_1(\cos\theta + i\sin\theta)=a+ib$ and $z_2=r_2(\cos\gamma+i\sin\gamma)=c+id$

Then, we have $\arg(z_1)=tan^{-1}(\frac{b}{a})$ and $\arg(z_2)=tan^{-1}(\frac{d}{c})$

As $\arg(z_1)=\arg(z_2)$ then $\tan^{-1}(\frac{b}{a})=\tan^{-1}(\frac{d}{c})\implies \frac{b}{a}=\frac{d}{c}\implies b.c=a.d$

This is correct, i have serious doubt about the interpretation of this. Can someone help me?

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  • $\begingroup$ You know the argument can be determined just up to an integer multiple of $\;2\pi\;$ radians, so... $\endgroup$ – DonAntonio Oct 13 '18 at 16:22
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It is known that $\arg\left(\frac{z_1}{z_2}\right)=\arg(z_1)-\arg(z_2)$, so this is equal to zero as $\arg(z_1)=\arg(z_2)$; so as $\frac{z_1}{z_2}$ has argument 0, it is a positive real number, so $z_1=kz_2$ for some positive real $k$.

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  • $\begingroup$ Oh.... Okay, now i understand. Thanks for all! $\endgroup$ – Bvss12 Oct 13 '18 at 16:34
  • $\begingroup$ One question, if $\frac{z_1}{z_2}$ have argument $0$ then $\frac{z_1}{z_2}=\frac{r_1}{r_2}\cos(\theta_1-\theta_2)$ no? then $k=\frac{r_1}{r_2}\cos(\theta_1-\theta_2)$ $\endgroup$ – Bvss12 Oct 13 '18 at 16:40
  • $\begingroup$ Yes, and $\theta_1-\theta_2=0$. $\endgroup$ – Benedict Randall Shaw Oct 13 '18 at 16:41
  • $\begingroup$ $\theta_1 - \theta_2$ is a multiple of $2\pi$ and $\cos(2n\pi) = 0$. $\endgroup$ – Trevor Gunn Oct 13 '18 at 16:45
  • $\begingroup$ @Trevor Gunn It was funny because until now I thought $\cos (2n\pi)=1$ ..... :-) Just kidding, might be a typo!! Cheers!!! $\endgroup$ – Rohan Shinde Oct 13 '18 at 16:54
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The whole purpose of the argument function is that if $z = re^{i\theta}$ then we have the absolute value function for $r$ (that is, $r = |z|$) and the argument function gives us $\theta = \arg(z)$ modulo a multiple of $2\pi$. But in any case we have

$$ e^{i\theta} = e^{i\arg(z)}. $$

Thus $z = |z|e^{i \arg(z)}$.

The purpose of writing the argument function in terms of the arctangent is to convert from cartesian coordinates ($a + bi$) to polar coordinates. But it's not exactly $\arctan(b/a)$, it's

$$ \arg(a + bi) = \arctan(a,b) = \begin{cases} \arctan(b/a) & \text{if } (a,b) \in \text{ quadrants I or IV} \\ \arctan(b/a) + \pi & \text{if } (a,b) \in \text{ quadrants II or III} \\ \end{cases} $$

Because, for example, $\arctan(\frac{-1}{-1}) = \arctan{\frac{1}{1}} = \frac{\pi}4$ even though the points $-1 - i$ and $1 + i$ are 180 degrees apart.

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A good way to see it is that $\cos\theta+i\sin\theta=\cos\gamma+i\sin\gamma$. So, if you put $k=r_2/r_1$, you get that $z_2=kz_1$: $$ z_2=r_2 (\cos\theta+i\theta)=kr_1(\cos\theta+i\theta)=kz_1. $$

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  • $\begingroup$ I don't understand... Why you say $z_1=z_2$? $\endgroup$ – Bvss12 Oct 13 '18 at 16:29
  • $\begingroup$ I don't. There's a $k $. $\endgroup$ – Martin Argerami Oct 13 '18 at 16:46

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