• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Complex Number Question (1 Viewer)

let.me.die

New Member
Joined
Sep 27, 2004
Messages
28
Gender
Undisclosed
HSC
2006
hi.. i just started complex numbers and i was wondering if someone could help me with some questions.
The points P and Q are represented by the complex numbers z = 1-3i and w = -3+4i respectively. FInd a point R on the real azis such that PRQ is a right-angled triangle.
 

Porcia

Member
Joined
Dec 27, 2004
Messages
256
Gender
Male
HSC
2006
no bloody idea, actually i'd like to find out too... seems like we're doing same question
 

Raginsheep

Active Member
Joined
Jun 14, 2004
Messages
1,227
Gender
Male
HSC
2005
Umm...let z be represented by P(1,-3) and w by Q(-3,4). Call R (x,0).

Use gradient formula to find gradient of PR and QR. Gradient PR x Gradient QR = -1. Then, solve for x.
 

let.me.die

New Member
Joined
Sep 27, 2004
Messages
28
Gender
Undisclosed
HSC
2006
thanx. i was wondering if i could just use co-ordinate geometry to find the answer. i got other questions if someone wouldnt mind answering them..
1. If |z| = |w|, prove that (z+w)/(z-w) is purely imaginary. By drawing a suitable diagram, give a geometrical interpretation of the result.

2. Prove that for any two complex numbers z1 and z2 |z1 + z2| more than or equal to |z1| - |z2|, assuming |z1| > |z2|. When does the equality sign hold?
 

Raginsheep

Active Member
Joined
Jun 14, 2004
Messages
1,227
Gender
Male
HSC
2005
1. Multiply top and bottom by the conjugate of (z-w). The bottom becomes |z-w|^2 which is real.

For the top, its (z+w)(z-w). Assume the underline means conjugate.
(z+w)(z-w)
=(z+w)(z-w)
=zz-ww-zw+zw
=|z|^2-|w|^2-zw+zw
But, |z|=|w|
=zw-zw
let z=(x1, y1), w=(x2, y2) and sub in.
Now expand and simplify and the real terms should disappear leaving only the imaginary.

Since top is imaginay and bottom is real, the whole thing is imaginary.
 

robbo_145

Member
Joined
Oct 19, 2004
Messages
80
Location
Scenic Central Coast
Gender
Male
HSC
2005
let.me.die said:
thanx. i was wondering if i could just use co-ordinate geometry to find the answer. i got other questions if someone wouldnt mind answering them..
1. If |z| = |w|, prove that (z+w)/(z-w) is purely imaginary. By drawing a suitable diagram, give a geometrical interpretation of the result.
drawing a diagram it can be seen that z+w and z-w are the diagonals of a rhombus of sides, hence they cross at right angles
i assume you've done De Moivre's thereom

let z+w = rCisΘ
z-w= qCis(Θ-π/2) //q,r are real
(z+w)/(z-w)= r/qCis(Θ-(Θ-π/2)) by De Moivre
(z+w)/(z-w) = r/qCis(π/2) which is purely imaginary
 

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

Top