Find the greatest value of argument- complex numbers

[chwala]

Homework Statement

Sketch the region that satisfies the inequalities $|z|≤3$ and $Im z≥2$

hence find the greatest value of arg $z$ for points in this region in radians.

Relevant Equations

complex numbers

since $|z|≤3$ →$z=0+0i$, therefore we shall have centre$(0,0)$ and radius $3$,
find my sketch below,

 

[chwala]

first let me ask is the shading of the region correct?

 

[etotheipi]

The sketch is good. So now, which point in the shaded region has the greatest argument? Determine the angle by drawing a triangle.

 

[chwala]

 

[chwala]

will the angle $m$ give us the greatest argument, i think i have to make use of two congruent right angles...right?

 

[etotheipi]

It'd be easier to just find the angle in the triangle you drew on the left, and then add $\pi/2$. Anyway, you know the hypotenuse (radius) and the vertical side, so you should be good to go.

 

[chwala]

It'd be easier to just find the angle in the triangle you drew on the left, and then add $\pi/2$. Anyway, you know the hypotenuse (radius) and the vertical side, so you should be good to go.

the radius would be $3$ what of the vertical side...why can't i see it aaargh??...or by accurate drawing, i should be able to get the values...

 

[etotheipi]

the radius would be $3$ what of the vertical side...why can't i see it aaargh??...or by accurate drawing, i should be able to get the values...

It's 2, right? It coincides with your line $Im(z) = 2$!

 

[chwala]

bingo boss

 

[chwala]

Just checking on this,
the least argument (in degrees) = $$tan ∅ = \frac {2}{2.23}=41.88^0$$
and largest argument (in degrees)= $$cos ∅ =\frac {2}{3}=48.19^0+90^0 = 138.19^0$$

 

[Office_Shredder]

A nice check here is if it was $Im(z)=3/\sqrt{2}\approx 2.12$, then the argument would be 135 degrees. Since it's a little smaller than that, you have to rotate down the circle towards the x axis, and expect to get a slightly larger number.

 

[chwala]

I am not getting what you are saying, to find argument, we move in an anticlockwise manner from the start point $(0,0)$...and my reference is on the second quadrant where have a right angle and i considered the sides, $3,2$ and $\sqrt 5$ to find the acute angle...I thought $Im Z≥2$aaaaaaaaargh thanks Office...i now get your view. You were giving me a quick way of checking...happy new year mate

 

[Office_Shredder]

I am not getting what you are saying, to find argument, we move in an anticlockwise manner from the start point $(0,0)$...and my reference is on the second quadrant where have a right angle and i considered the sides, $3,2$ and $\sqrt 5$ to find the acute angle...

I think your answer is right, I'm just describing a way you can double check that you got at least close to the right answer (for the largest argument)