## Calculator Group Title 7ii) Complex analysis. Argand diagram http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_s10_qp_31.pdf 2 years ago 2 years ago

1. Calculator

check |dw:1348800828515:dw| how do you draw the inequalities?

2. Calculator

@UnkleRhaukus

3. vf321

2+2i will be at angle pi/2, but has a distance from the origin of $$2\sqrt 2$$, not 1.

4. Calculator

|dw:1348803360924:dw|

5. vf321

correct, but not to scale

6. vf321

im working on the ineqs btw, just wait plz

7. Calculator

k

8. vf321

Okay. got it.

9. Calculator

can you draw it out?

10. vf321

Yes. But ill just give u the eq., and you'll draw it.

11. vf321

Denote imaginary number z as a vector (a, b) where a, b are reals. Note that this means that on Argand Diagram x = a and y = b. Note also that |z| = |(a,b)| = sqrt(a^2+b^2). Then since $|z-1|\le|z-i|$$|z-1|^2\le|z-i|^2$$|(a-1, b)|^2\le|(a, b-1)|^2$$(a-1)^2+b^2\le a^2+(b-1)^2$

12. vf321

Simplify that and you'll get a nice cartesian inequality

13. vf321

Follow similar steps for the other one.

14. UnkleRhaukus

|dw:1348805078839:dw|

15. Calculator

why is it pi/2? $\arg ,~z=\tan^{-1}(\frac{b}{a})=\tan^{-1}\frac{2}{2}=\frac{\pi}{4}$

16. UnkleRhaukus

oh, i ment that

17. UnkleRhaukus

|dw:1348805469963:dw|

18. Calculator

so how do you sketch the inequalities? i have to go now , can you leave some tips or answer later ill go through it. thanks!

19. UnkleRhaukus

hmm im not sure about the inequalities yet , ill think about them

20. Calculator

@Directrix

21. perl

how did you derive |z - 1 | < | z - i | ?

22. Directrix

We can't let this question pass without paying homage of Jean Robert Argand who was an amateur mathematician. His day job was that of an accountant. It is remarkable and inspirational that although this diagram is Argand's only contribution to mathematics, it is, nonetheless, a significant one. http://www-groups.dcs.st-and.ac.uk/history/Biographies/Argand.html

23. perl

see if you can solve the first question

24. perl

| x + 3a | > 2 | x - 2a | , a > 0 , solve

25. perl

i think you can do that in cases, case 1 : x > -3a

26. Calculator

for that you just square both sides and get answer

27. Calculator

$x^2+6ax+9a^2 > 4(x^2-4ax+4a^2)$

28. perl

ok

29. Calculator

$-3x^2+22ax-7>0$

30. perl

yes

31. perl

32. Calculator

$x=\frac{11}{3}a \pm\frac{\sqrt{547}}{3}$

33. perl

what about the other ones, can you solve them

34. myko

|dw:1348823976443:dw| hint for 7 ii

35. Directrix

Spoilers/Hints/Maybe from Mark Scheme for Exam http://www.xtremepapers.com/papers/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20(9709)/9709_s10_ms_31.pdf

36. Calculator

|dw:1348824437459:dw| u could be there

37. Calculator

|dw:1348824571707:dw|

38. Calculator

@amistre64

39. amistre64

im not to confident in what args are. Ive never really delved into the language for them to be sure that what i know is what i know :)

40. phi

|z-1| ≤ |z -i| if this were the equality |z-1| = |z -i| it can be interpreted as the points z equidistant from the points (1,0) and (0,1) on the complex plane. The locus of points equidistant from 2 points is the perpendicular bisector of the line joining the points. For the inequality |z-1| ≤ |z -i|, the locus is all points on and below the line defined by the points (0,0) and (1/2, 1/2) |z-u| ≤ 1 if an equality, the locus is the circle with radius 1 located at u for the inequality, it is the filled in circle. the intersection of the two constraints is the half circle with center 2+2i and radius 1, (the half region below the line going from (0,0) through (2,2)