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

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?

• You know the argument can be determined just up to an integer multiple of $\;2\pi\;$ radians, so... – DonAntonio Oct 13 '18 at 16:22

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 $$k$$.

• Oh.... Okay, now i understand. Thanks for all! – Bvss12 Oct 13 '18 at 16:34
• 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 $k=\frac{r_1}{r_2}\cos(\theta_1-\theta_2)$ – Bvss12 Oct 13 '18 at 16:40
• Yes, and $\theta_1-\theta_2=0$. – Benedict Randall Shaw Oct 13 '18 at 16:41
• $\theta_1 - \theta_2$ is a multiple of $2\pi$ and $\cos(2n\pi) = 0$. – Trevor Gunn Oct 13 '18 at 16:45
• @Trevor Gunn It was funny because until now I thought $\cos (2n\pi)=1$ ..... :-) Just kidding, might be a typo!! Cheers!!! – Rohan Shinde Oct 13 '18 at 16:54

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.

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 $$z_2=kz_1$$: $$z_2=r_2 (\cos\theta+i\theta)=kr_1(\cos\theta+i\theta)=kz_1.$$

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