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$$S=\{z ∈ C : |z + \frac{13i}{(3-2i)}| < 2 ∧ -\frac{2π}{3} ≤ Arg(7z) ≤ -\frac{π}{6}\}$$

I multiplied $\frac{13i}{(3-2i)}$ by $3+2i$, so that's what i got $$S=\{z ∈ C : |z + (-2+3i)| < 2 ∧ -\frac{2π}{3} ≤ Arg(7z) ≤ -\frac{π}{6}\}$$ but I don't understand what's next. We know that $z$ is complex number, it means that $z = x + iy$, but what can we do with it? here

Andrii Dovhun
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    I think that, at some point, you need to start translating the algebra into geometry. You do, after all, need a geometric answer at the end. Why not start now? What is the geometric interpretation of each of the two conditions you have there? – Arthur Apr 24 '23 at 07:12
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    Hint ${z\mid |z-a|<b}$ is the open ball of center $a$ and radius $b$. Moreover, $\arg(7z)=\arg(z)$... – Surb Apr 24 '23 at 07:13
  • @Surb, I think I understood your hint, but I didn't get the logic behind it. Can you explain, please, why arg(7)=arg() – Andrii Dovhun Apr 24 '23 at 07:26
  • @AndriiDovhun I think you should look into the geometry of that statement. What does it say about points in the plane and angles? – Arthur Apr 24 '23 at 07:38
  • Let $~x + iy~ = z,~$ and let $~\displaystyle r = |z| = \sqrt{x^2 + y^2}.$ Then Arg$(z)~$ is the unique angle $~\theta,~$ within a modulus of $~(2\pi),~$ such that $$\cos(\theta) = \frac{x}{r}, ~\sin(\theta) = \frac{y}{r}.$$ Now, let $~w = 7z = 7x + i7y.~$ Let $~R~$ denote $~|w| \implies R = 7r.~$ Then, Arg$(w)~$ is the unique angle $~\alpha,~$ within a modulus of $~2\pi,~$ such that $$\cos(\alpha) = \frac{7x}{R}, ~\sin(\alpha) = \frac{7y}{R}.$$ So, $$\cos(\alpha) = \cos(\theta), ~\sin(\alpha) = \sin(\theta) \implies \alpha = \theta.$$ – user2661923 Apr 24 '23 at 08:18
  • @user376343: Obviously, in the context of the exercise, $\arg(z)\in [-\pi,\pi)$ and thus, $\arg(7)=0...$ ;-) – Surb Apr 24 '23 at 09:50
  • @Surb I wrongly added your name instead of OP. I wanted explain them that you're right. My apologies. – user376343 Apr 24 '23 at 15:21

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