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

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- katieb

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

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@UnkleRhaukus

- anonymous

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

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|dw:1348803360924:dw|

- anonymous

correct, but not to scale

- anonymous

im working on the ineqs btw, just wait plz

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k

- anonymous

Okay. got it.

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can you draw it out?

- anonymous

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

- anonymous

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
\[|z-1|\le|z-i|\]\[|z-1|^2\le|z-i|^2\]\[|(a-1, b)|^2\le|(a, b-1)|^2\]\[(a-1)^2+b^2\le a^2+(b-1)^2\]

- anonymous

Simplify that and you'll get a nice cartesian inequality

- anonymous

Follow similar steps for the other one.

- UnkleRhaukus

|dw:1348805078839:dw|

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

- UnkleRhaukus

oh, i ment that

- UnkleRhaukus

|dw:1348805469963:dw|

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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!

- UnkleRhaukus

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

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@Directrix

- perl

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

- 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

- perl

see if you can solve the first question

- perl

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

- perl

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

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for that you just square both sides and get answer

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\[x^2+6ax+9a^2 > 4(x^2-4ax+4a^2)\]

- perl

ok

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\[-3x^2+22ax-7>0\]

- perl

yes

- perl

then use the quadratic formula?

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

- perl

what about the other ones, can you solve them

- anonymous

|dw:1348823976443:dw|
hint for 7 ii

- 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

##### 1 Attachment

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|dw:1348824437459:dw|
u could be there

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|dw:1348824571707:dw|

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@amistre64

- 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 :)

- 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)

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