Complex Number question help!! (1 Viewer)

ledamn

New Member
Joined
Sep 18, 2014
Messages
20
Gender
Male
HSC
2015
If anyone would like to try to solve these questions, then you're welcome!

2) If |z| = 1, where a is a positive number, show that:
a) arg(z+1) = 1/2argz Easy(done)
b) |z+1| = 2sin(arg(z-1))

6) On the Argand diagram, the points P, Q, and R represent the complex number p, q, & r respectively. If pq +iq +ir = 0, what triangle is PQR?

7) Find all possible values for z4 if z1= 1+i, z2=2+6i, z3=-1+7i and z4 forms a parallelogram.


Thanks everybody!
 

Triage

Member
Joined
May 30, 2012
Messages
245
Gender
Undisclosed
HSC
N/A
If anyone would like to try to solve these questions, then you're welcome!

2) If |z| = 1, where a is a positive number, show that:
a) arg(z+1) = 1/2argz Easy(done)
b) |z+1| = 2sin(arg(z-1))

6) On the Argand diagram, the points P, Q, and R represent the complex number p, q, & r respectively. If pq +iq +ir = 0, what triangle is PQR?

7) Find all possible values for z4 if z1= 1+i, z2=2+6i, z3=-1+7i and z4 forms a parallelogram.


Thanks everybody!
2B

RHS = 2sin(tan^-1(Y/X-1)

RHS^2 = 4sin^2(tan^-1(Y/X-1)

= 4 ( tan^2(tan^-1(Y/X-1)) * cos^2(tan^-1(Y/X-1)))

= 4((Y/X-1)^2 )* (1/1+(Y/X-1)^2))
 

Triage

Member
Joined
May 30, 2012
Messages
245
Gender
Undisclosed
HSC
N/A
I don't understand what you did?
Nah I think I might be wrong.

Lets let Z = X + iY

therefor Z + 1 = X+1 + iY

yeah yeah yeah I see my mistake now.

so RHS^2 will equal 4sin squared of inverse tan of y/x+1

now sin squared can be re-written as cos squared + tan squared

thus tan^2 of the inverse tan of y/x+1 all multiplied by cos squared of the inverse tan of y/x+1

now cos^2 = 1/1+tan^2
 

Axio

=o
Joined
Mar 20, 2014
Messages
484
Gender
Male
HSC
2015
If anyone would like to try to solve these questions, then you're welcome!
6) On the Argand diagram, the points P, Q, and R represent the complex number p, q, & r respectively. If pq +iq +ir = 0, what triangle is PQR?
Thanks everybody!
Ok yeah I was wrong, I was doing it for pq + pr +qr = 0
 
Last edited:
Joined
Sep 29, 2013
Messages
196
Gender
Male
HSC
2016
Nope, it's wrong. Question 6 is an isosceles right angled triangle at q. But I don't know how to show working out though
 

ledamn

New Member
Joined
Sep 18, 2014
Messages
20
Gender
Male
HSC
2015
Yeppe @Triage, I definitely think to get the solutions for 2B, you need the Argand Diagram on 2A.
Also quesiton 6 is difficult, don't know the answers tbh.
 

actuallyapotato

New Member
Joined
Aug 19, 2014
Messages
1
Gender
Undisclosed
HSC
2014
Is it too late to post a reply? I'm new and I don't know how long the appropriate timeframe to reply is.

I found 2B by using this Argand diagram

Untitled.png

Then after messing with it a bit you'll find the angle between the lines representing z-1 and z+1 is 90 degrees, and you get the solution. :)
 

ledamn

New Member
Joined
Sep 18, 2014
Messages
20
Gender
Male
HSC
2015
Ah bro, that's still wrong LOL. I got the answer for every problem on top. If you guys want me to post it?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top