Complex Numbers Questions (1 Viewer)

[lychnobity]

Hey guys, can't do any of these questions, so I'm hoping that you'll help me out.


3) P & Q are 2 points representing complex numbers z₁ and z₂ respectively in the complex number plane such that z₁ - z₂ = 2i(z₁ + z₂). O is the origin.

a) By writing z₁ - z₂ = 2i(z₁ + z₂) in vectors, explain why OP = OQ

b) Show that tan (angle POQ) = -4/3


6a) If Re(z² + z) is a constant find the locus of z.
b) If Re(z) is a constant find the locus of z² + z

 

[untouchablecuz]

Hey guys, can't do any of these questions, so I'm hoping that you'll help me out.


3) P & Q are 2 points representing complex numbers z₁ and z₂ respectively in the complex number plane such that z₁ - z₂ = 2i(z₁ + z₂). O is the origin.

a) By writing z₁ - z₂ = 2i(z₁ + z₂) in vectors, explain why OP = OQ

b) Show that tan (angle POQ) = -4/3


6a) If Re(z² + z) is a constant find the locus of z.
b) If Re(z) is a constant find the locus of z² + z

3. a)

let P represent z₁ and Q represent z₂

by addition OC represents z₁ + z₂

by subtraction PQ represents z₁ - z₂

but it is stipulated that z₁ - z₂ = 2i(z₁ + z₂)

hence PQ = 2i(z₁ + z₂)

thus OC/PQ = (1/2i) => (2i)OC = PQ

hence PQ is a rotation of OC by pi/2 followed by an enlargement by a factor of 2

thus the diagonals intersect at right angles, implying that OPCQ is a rhombus

thus OP = OQ

(dont forget the little arrows over ther "OP" etc)

b) let the intersection between the diagonals be D

let DOQ = x

since the diagonals in a rhombus bisect the angles,

our required angle POQ = 2x

now consider the RIGHT triangle ODQ

tan(x) = DQ/DO

But PQ and OC are in the ratio 2:1

thus DQ/OD = (2/2)/(1/2) (diagonals bisect one another) => tan(x) = 2

tan(POQ) = tan2x = 2tanx/(1- tan^2x) = (2.2)/(1-2^2) = -4/3

6 a) i assume all you do is sub z = x +iy into z² + z and find the real part, and let it equal a constant, such as k and i think you end up with a rectangular hyperbola

b) not 100% sure

 

[Affinity]

let z = a+ib,

(a^2 + a - b^2) + i(2a+1)b

if a is not -1/2 then
y^2 / (2a+1)^2 + x = a^2 + a
gives a parabola.

for the special case Re(z) = a = -1/2, you get a ray on the x-axis.

 

[untouchablecuz]

let z = a+ib,

(a^2 + a - b^2) + i(2a+1)b

if a is not -1/2 then
y^2 / (2a+1)^2 + x = a^2 + a
gives a parabola.

for the special case Re(z) = a = -1/2, you get a ray on the x-axis.

is this for 6 b)?

hmm, i don't understand; was i correct in my method for 6 a)?

also, what's exactly classified as a constant? is the constant a real/imaginary/complex number or either one?