2
$\begingroup$

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:

  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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Viktor Glombik Oct 28 at 17:23
  • $\begingroup$ @ViktorGlombik I edited it. $\endgroup$ – l0ner9 Oct 28 at 17:46
-1
$\begingroup$

By definition $\frac{π}3≤\arg(z)≤π$ represents all complex numbers contained in the first and second quadrant between the lines

  • $y=\tan (\pi/3) \cdot x$
  • $y=0$
$\endgroup$
  • $\begingroup$ Could you elaborate a little further? What does that represent on the complex plane? $\endgroup$ – l0ner9 Oct 28 at 17:55
  • $\begingroup$ @l0ner9 The condition represents the points in between line $y=\tan (\pi/3) \cdot x =\sqrt 3 x$ and the negative $x$ axis. $\endgroup$ – user Oct 28 at 18:01
  • $\begingroup$ 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? $\endgroup$ – l0ner9 Oct 28 at 18:25
  • $\begingroup$ 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. $\endgroup$ – user Oct 28 at 19:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.