Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
 one year ago
 one year ago
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
 one year ago
 one year ago

This Question is Closed

Calculator Group TitleBest ResponseYou've already chosen the best response.0
check dw:1348800828515:dw how do you draw the inequalities?
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
@UnkleRhaukus
 one year ago

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

Calculator Group TitleBest ResponseYou've already chosen the best response.0
dw:1348803360924:dw
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
correct, but not to scale
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
im working on the ineqs btw, just wait plz
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Okay. got it.
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
can you draw it out?
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Yes. But ill just give u the eq., and you'll draw it.
 one year ago

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

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Simplify that and you'll get a nice cartesian inequality
 one year ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Follow similar steps for the other one.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1348805078839:dw
 one year ago

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
oh, i ment that
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1348805469963:dw
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
hmm im not sure about the inequalities yet , ill think about them
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
@Directrix
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
how did you derive z  1  <  z  i  ?
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.0
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://wwwgroups.dcs.stand.ac.uk/history/Biographies/Argand.html
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
see if you can solve the first question
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
 x + 3a  > 2  x  2a  , a > 0 , solve
 one year ago

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

Calculator Group TitleBest ResponseYou've already chosen the best response.0
for that you just square both sides and get answer
 one year ago

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

Calculator Group TitleBest ResponseYou've already chosen the best response.0
\[3x^2+22ax7>0\]
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
then use the quadratic formula?
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
\[x=\frac{11}{3}a \pm\frac{\sqrt{547}}{3}\]
 one year ago

perl Group TitleBest ResponseYou've already chosen the best response.0
what about the other ones, can you solve them
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.0
dw:1348823976443:dw hint for 7 ii
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
dw:1348824437459:dw u could be there
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
dw:1348824571707:dw
 one year ago

Calculator Group TitleBest ResponseYou've already chosen the best response.0
@amistre64
 one year ago

amistre64 Group TitleBest ResponseYou've already chosen the best response.0
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 :)
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
z1 ≤ 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)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.