# Double inequality containing the argument of a complex number and two angle values

I have come across an exercise which requires me to plot the area which satisfies these two conditions:

$$|z|^2 -5|z| +6<0$$

$$\frac{π}3≤arg(z)≤π$$

Now, I know how to solve the first one but the second one baffles my mind.

Correct me where I'm wrong:

$$arg(z)$$ is typically defined as $$arctan \frac{y}x$$

If I apply it to the double inequality above, I get $$\frac{π}3≤arctan\frac{y}x≤π$$

Now, if we divide it into two inequalities and perform the tangent operation on both sides we get:

$$\frac{y}x ≥ \sqrt{3}$$ and $$\frac{y}x ≤ 0$$

Which of course has no solutions. However, looking at the inequality, I have a notion that (graphically) the solutions are all angles between π and $$\frac{π}3$$, going clockwise. However, I have no idea how to solve this. Can anyone help?

Note: I am making elementary errors and please be gentle as this is the first time I'm encountering these kinds of inequalities.

EDIT: As requested, providing the solution to the first equation:

Let $$|z| = t$$

$$t^2 -5t +6 <0$$, therefore

$$(t-3)(t-2)<0$$

$$(t<3 \land t>2) \lor (t>3 \land t<2)$$

Substituting t and squaring we get:

1. $$\sqrt{(x^2+y^2)} <3^2 \land \sqrt{(x^2+y^2)}>2^2$$

Graphically this would represent the annulus represented by the two circles. The other set of equations has no solution.

• Why would the be no solutions? The first inequality is equivalent to $y \ge \sqrt{3}x$ and the second to $y \le 0$ and $x\ne0$. This is fulfilled for example for $x = y= -1$ as $-1 \ge - \sqrt{3}$ is equivalent to $1\le \sqrt{3}$. Also, your graphic notion is correct, since the argument is precisely the angle a vector connecting the origin to the complex number makes with the real axis. Please show how you have solved the first one. – Viktor Glombik Oct 28 at 17:23
• @ViktorGlombik I edited it. – l0ner9 Oct 28 at 17:46

By definition $$\frac{π}3≤\arg(z)≤π$$ represents all complex numbers contained in the first and second quadrant between the lines
• $$y=\tan (\pi/3) \cdot x$$
• $$y=0$$
• @l0ner9 The condition represents the points in between line $y=\tan (\pi/3) \cdot x =\sqrt 3 x$ and the negative $x$ axis. – user Oct 28 at 18:01
• Thanks! I have one more question: Why is it the I and II quadrants? There are points in between the two lines but which are in the III and IV quadrants. And also, isn't $\frac{y}x ≤0$ the second and fourth quadrant? – l0ner9 Oct 28 at 18:25
• Because a certain value for $\arg(z)$ indicates a ray and not a line. For example $\arg(z)=0$ represents the positive $x$ axis and $\arg(z)=\pi$ the negative $x$ axis. – user Oct 28 at 19:16