A task about argument

[ricsi046]

Hello,Can i get some help with this task?
I need to illustrate those z complex numbers on the complex plane for which:
pi/2 <= arg(z+1/z) <= pi
I have tried everything i could,but no result so far.Should i write up something like this?Let's say that z=a+bi,then (z+1/z)=a+bi+(a-bi/a²+b²),so
b <= a+bi+(a-bi/a²+b²) <= a
Or should i use trigonometric form?

 

[pka]

illustrate those z complex numbers on the complex plane for which:
pi/2 <= arg(z+1/z) <= pi
z=a+bi,then (z+1/z)=a+bi+(a-bi/a²+b²),so
a+bi+(a-bi/a²+b²) CORRECT!

In order for $\displaystyle \dfrac{\pi }{2} \le \arg (z) \le \pi$ you must have $\displaystyle \text{Re}(z)\le 0$
and $\displaystyle \text{Im}(z)\ge 0$.

Because $\displaystyle \left( {z + {z^{ - 1}}} \right) = \left( {\dfrac{{{{\left| z \right|}^2}z + \overline z }}{{{{\left| z \right|}^2}}}} \right)$ you can see that
$\displaystyle \text{Re}\left( {z + {z^{ - 1}}} \right)\le 0$ if $\displaystyle \text{Re}(z)\le 0$ and

$\displaystyle \text{Im}\left( {z + {z^{ - 1}}} \right)\ge 0$ if $\displaystyle (|z|^2-1)\text{Im}(z)\ge 0.$

Your graph is in the left half-plane. In the upper half-plane it is outside the circle $\displaystyle |z|-1$
In the lower half-plane it is inside the circle $\displaystyle |z|-1$.

 

[ricsi046]

In order for $\displaystyle \dfrac{\pi }{2} \le \arg (z) \le \pi$ you must have $\displaystyle \text{Re}(z)\le 0$
and $\displaystyle \text{Im}(z)\ge 0$.

Because $\displaystyle \left( {z + {z^{ - 1}}} \right) = \left( {\dfrac{{{{\left| z \right|}^2}z + \overline z }}{{{{\left| z \right|}^2}}}} \right)$ you can see that
$\displaystyle \text{Re}\left( {z + {z^{ - 1}}} \right)\le 0$ if $\displaystyle \text{Re}(z)\le 0$ and

$\displaystyle \text{Im}\left( {z + {z^{ - 1}}} \right)\ge 0$ if $\displaystyle (|z|^2-1)\text{Im}(z)\ge 0.$

Your graph is in the left half-plane. In the upper half-plane it is outside the circle $\displaystyle |z|-1$
In the lower half-plane it is inside the circle $\displaystyle |z|-1$.

thank you so much The idea that re()<=0 and im()>=0 is what i needed,from there it was just calculation with a's and b's
I made a paint picture just to make sure that it's correct.I want to ask if the points on the circle are included too, right?Both under and above the real axis

 

[pka]

I made a paint picture just to make sure that it's correct.I want to ask if the points on the circle are included too, right?Both under and above the real axis

You are correct. Yes the points on the circle are included.