Thanks for taking your time to read this question.
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:
- $\sqrt{(x^2+y^2)} <3^2 \land \sqrt{(x^2+y^2)}>2^2$
Graphically this would represent the annulus represented by the two circles. The other set of equations has no solution.