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7ii) Complex analysis. Argand diagram

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check |dw:1348800828515:dw| how do you draw the inequalities?
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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correct, but not to scale
im working on the ineqs btw, just wait plz
Okay. got it.
can you draw it out?
Yes. But ill just give u the eq., and you'll draw it.
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 \[|z-1|\le|z-i|\]\[|z-1|^2\le|z-i|^2\]\[|(a-1, b)|^2\le|(a, b-1)|^2\]\[(a-1)^2+b^2\le a^2+(b-1)^2\]
Simplify that and you'll get a nice cartesian inequality
Follow similar steps for the other one.
why is it pi/2? \[\arg ,~z=\tan^{-1}(\frac{b}{a})=\tan^{-1}\frac{2}{2}=\frac{\pi}{4}\]
oh, i ment that
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!
hmm im not sure about the inequalities yet , ill think about them
how did you derive |z - 1 | < | z - i | ?
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.
see if you can solve the first question
| x + 3a | > 2 | x - 2a | , a > 0 , solve
i think you can do that in cases, case 1 : x > -3a
for that you just square both sides and get answer
\[x^2+6ax+9a^2 > 4(x^2-4ax+4a^2)\]
then use the quadratic formula?
\[x=\frac{11}{3}a \pm\frac{\sqrt{547}}{3}\]
what about the other ones, can you solve them
|dw:1348823976443:dw| hint for 7 ii
Spoilers/Hints/Maybe from Mark Scheme for Exam
1 Attachment
|dw:1348824437459:dw| u could be there
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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