rosemariechua
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(i) On a sketch of an Argand diagram, shade the region whose points represent complex
numbers satisfying the inequalities Ï − 2 − i ≤ 1 and Ï − i ≤ Ï − 2. [4]
(ii) Calculate the maximum value of arg Ï for points lying in the shaded region.


https://pastpapers.co/cie/view.php?i..._s14_qp_33.pdf


Can someone help me please? IDK how too draw .....
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begbie68
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you should be able to find (easily) youtubes & other web resources explaining loci in the complex plane.
It's also a standard topic for Further Maths topics all boards (especially Edexcel)

|z -(a+ib)| = r will give you a circle centre (a,b) and radius r
|z -(a+ib)| < r are the points strictly inside the circle ( with the > instead, you'll get the pts OUTside the circle)

|z-(a+ib)| = |z-(c+id)| will give you an infinitely long straight line which is the same as the perpendicular bisector of the pts (a,b) & (c,d)
|z-(a+ib)| < |z-(c+id)| will give you all the pts closer to (a,b) than (c,d) - in other words the shaded part including (a,b) to one side of your pependicular bisector.

The you tubes & text books will give you the full explanation & proofs/evidence for these.

hth
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rosemariechua
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Thank you so much for the helpp!!
(Original post by begbie68)
you should be able to find (easily) youtubes & other web resources explaining loci in the complex plane.
It's also a standard topic for Further Maths topics all boards (especially Edexcel)

|z -(a+ib)| = r will give you a circle centre (a,b) and radius r
|z -(a+ib)| < r are the points strictly inside the circle ( with the > instead, you'll get the pts OUTside the circle)

|z-(a+ib)| = |z-(c+id)| will give you an infinitely long straight line which is the same as the perpendicular bisector of the pts (a,b) & (c,d)
|z-(a+ib)| < |z-(c+id)| will give you all the pts closer to (a,b) than (c,d) - in other words the shaded part including (a,b) to one side of your pependicular bisector.

The you tubes & text books will give you the full explanation & proofs/evidence for these.

hth
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rosemariechua
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Hiii can help me with these questions too please? Kinda struggling ..
Ans Q5) i) 12ms
Ii!) 2ms^2
iii) 13.6ms
Q7)i)AB 6.4N
BC 4.8N
ii) 0.359
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rosemariechua
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(Original post by begbie68)
you should be able to find (easily) youtubes & other web resources explaining loci in the complex plane.
It's also a standard topic for Further Maths topics all boards (especially Edexcel)

|z -(a+ib)| = r will give you a circle centre (a,b) and radius r
|z -(a+ib)| < r are the points strictly inside the circle ( with the > instead, you'll get the pts OUTside the circle)

|z-(a+ib)| = |z-(c+id)| will give you an infinitely long straight line which is the same as the perpendicular bisector of the pts (a,b) & (c,d)
|z-(a+ib)| < |z-(c+id)| will give you all the pts closer to (a,b) than (c,d) - in other words the shaded part including (a,b) to one side of your pependicular bisector.

The you tubes & text books will give you the full explanation & proofs/evidence for these.

hth
hello how to find the maximum value of argz?
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old_engineer
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(Original post by rosemariechua)
Hiii can help me with these questions too please? Kinda struggling ..
Ans Q5) i) 12ms
Ii!) 2ms^2
iii) 13.6ms
Q7)i)AB 6.4N
BC 4.8N
ii) 0.359
5 i) P loses PE and gains the same amount of KE.
5 ii) Fnet = ma in the direction parallel with the slope.
5 iii) SUVAT

7 i) Resolve horizontally and vertically at B. Net force is zero in both directions.
7 ii) Resolve forces horizontally and vertically at A, then use F = (mu)R
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rosemariechua
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(Original post by old_engineer)
5 i) P loses PE and gains the same amount of KE.
5 ii) Fnet = ma in the direction parallel with the slope.
5 iii) SUVAT

7 i) Resolve horizontally and vertically at B. Net force is zero in both directions.
7 ii) Resolve forces horizontally and vertically at A, then use F = (mu)R
hello for 7i) I dont get it .....
what I did: 8sin(53.13)=6.4
however its for BC and not for AB .. why is that? can explain please?
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old_engineer
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(Original post by rosemariechua)
hello for 7i) I dont get it .....
what I did: 8sin(53.13)=6.4
however its for BC and not for AB .. why is that? can explain please?
Not sure what you're trying to do there, but, because ABC is a right angled triangle, sinA = cosC, and cosA = sinC. You shouldn't need to refer to the angles in degrees as the relevant sine and cosine values will come from the three side lengths.

In the vertical direction, Tc.cosC = Ta.cosA. Now can you construct a similar equation for the horizontal direction?
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