arg diagrams (1 Viewer)

[kkk579]

for this graph it was to shade -pi/2 <= arg <= - pi then the bottom greater part would be shaded right

 

[kkk579]

For this q shouldnt the angle be theta - pi/3 ?

 

[kkk579]

For the cambridge q, it says that the range of arg(Z) is the same cos its just z being shifted but howcome the same logic doesnt apply to the other q i attached? I understand how the fort st q arrives to the answer but im js confused because the 2 solutions seem a bit contradictory

 

[fluffchuck]

for this graph it was to shade -pi/2 <= arg <= - pi then the bottom greater part would be shaded rightView attachment 45867View attachment 45868

The idea behind your region shading is correct! Just be careful with how you position the lines for the arg locus & the circle for the modulus locus - hint: the arg locus stems from the point (1,1), does this point (1,1) lie inside the circle, or does it lie on the circle?

 

[fluffchuck]

View attachment 45869View attachment 45870
For this q shouldnt the angle be theta - pi/3 ?

The answer is correct here, the angle is = (big angle on diagram) - (small angle on diagram),
the big angle on the diagram is pi/3,
the small angle on the diagram is pi

I take this back, you are correct - the angle is indeed a negative angle, and it should be -[(big angle on diagram) - (small angle on diagram)] = (small angle on diagram) - (big angle on diagram)

 

[fluffchuck]

For the cambridge q, it says that the range of arg(Z) is the same cos its just z being shifted but howcome the same logic doesnt apply to the other q i attached? I understand how the fort st q arrives to the answer but im js confused because the 2 solutions seem a bit contradictoryView attachment 45871View attachment 45872View attachment 45873View attachment 45874

The reason why the Cambridge solution & the Fort St solution is a bit different is because:
- In the Cambridge question, part (b) shifts both the argument and the circle by 2 units to the left, i.e. |z-2|=1 becomes |z|=1, and arg(z) becomes arg(z+2). As a result, the diagram would look exactly like (a), but everything is just moved 2 units to the left.
- In the Fort st question, the circle remains stationary, and only the argument shifts, i.e. the question contains arg(z), arg(z-1), and arg(z-2). As such, the circle remains still, whereas the arguments are constantly moving relative to the circle.

Hope this helps!

 

[kkk579]

The reason why the Cambridge solution & the Fort St solution is a bit different is because:
- In the Cambridge question, part (b) shifts both the argument and the circle by 2 units to the left, i.e. |z-2|=1 becomes |z|=1, and arg(z) becomes arg(z+2). As a result, the diagram would look exactly like (a), but everything is just moved 2 units to the left.
- In the Fort st question, the circle remains stationary, and only the argument shifts, i.e. the question contains arg(z), arg(z-1), and arg(z-2). As such, the circle remains still, whereas the arguments are constantly moving relative to the circle.

Hope this helps!

thank you this helps alot!

 

[kkk579]

The idea behind your region shading is correct! Just be careful with how you position the lines for the arg locus & the circle for the modulus locus - hint: the arg locus stems from the point (1,1), does this point (1,1) lie inside the circle, or does it lie on the circle?

wait but why is it the outside part i thought it would be the larger bottom bit

 

[fluffchuck]

wait but why is it the outside part i thought it would be the larger bottom bit

Sorry I'm not sure if I'm following - what do you mean by this?

 

[kkk579]

Sorry I'm not sure if I'm following - what do you mean by this?

Like as in why do u shade the part i attached in the photo and not the region below the 2 perpendicular rays

 

[fluffchuck]

Like as in why do u shade the part i attached in the photo and not the region below the 2 perpendicular rays

It is because the question is looking for -pi/2 <= arg(z - 1 - i) <= pi, the bit that isn't shaded in your diagram represents -pi <= arg(z - 1 - i) <= -pi/2