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 2 years ago
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
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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Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0check dw:1348800828515:dw how do you draw the inequalities?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.12+2i will be at angle pi/2, but has a distance from the origin of \(2\sqrt 2\), not 1.

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1348803360924:dw

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1correct, but not to scale

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1im working on the ineqs btw, just wait plz

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0can you draw it out?

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Yes. But ill just give u the eq., and you'll draw it.

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Denote 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 \[z1\lezi\]\[z1^2\lezi^2\]\[(a1, b)^2\le(a, b1)^2\]\[(a1)^2+b^2\le a^2+(b1)^2\]

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Simplify that and you'll get a nice cartesian inequality

vf321
 2 years ago
Best ResponseYou've already chosen the best response.1Follow similar steps for the other one.

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1348805078839:dw

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0why is it pi/2? \[\arg ,~z=\tan^{1}(\frac{b}{a})=\tan^{1}\frac{2}{2}=\frac{\pi}{4}\]

UnkleRhaukus
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1348805469963:dw

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0so 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
 2 years ago
Best ResponseYou've already chosen the best response.1hmm im not sure about the inequalities yet , ill think about them

perl
 2 years ago
Best ResponseYou've already chosen the best response.0how did you derive z  1  <  z  i  ?

Directrix
 2 years ago
Best ResponseYou've already chosen the best response.0We 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://wwwgroups.dcs.stand.ac.uk/history/Biographies/Argand.html

perl
 2 years ago
Best ResponseYou've already chosen the best response.0see if you can solve the first question

perl
 2 years ago
Best ResponseYou've already chosen the best response.0 x + 3a  > 2  x  2a  , a > 0 , solve

perl
 2 years ago
Best ResponseYou've already chosen the best response.0i think you can do that in cases, case 1 : x > 3a

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0for that you just square both sides and get answer

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0\[x^2+6ax+9a^2 > 4(x^24ax+4a^2)\]

perl
 2 years ago
Best ResponseYou've already chosen the best response.0then use the quadratic formula?

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0\[x=\frac{11}{3}a \pm\frac{\sqrt{547}}{3}\]

perl
 2 years ago
Best ResponseYou've already chosen the best response.0what about the other ones, can you solve them

myko
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1348823976443:dw hint for 7 ii

Directrix
 2 years ago
Best ResponseYou've already chosen the best response.0Spoilers/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

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1348824437459:dw u could be there

Calculator
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1348824571707:dw

amistre64
 2 years ago
Best ResponseYou've already chosen the best response.0im 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
 2 years ago
Best ResponseYou've already chosen the best response.0z1 ≤ z i if this were the equality z1 = 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 z1 ≤ z i, the locus is all points on and below the line defined by the points (0,0) and (1/2, 1/2) zu ≤ 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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