More Complex No. Q's (1 Viewer)

[scardizzle]

http://i.imgur.com/BBY7H.jpg

Sorry for all these questions, I definitely need more practice with complex no.'s. Any help is greatly appreciated.

 

[dasicmankev]

For the first question, this method is a bit tedious and I'm sure there's an easier way, but here it goes:

Let z = x + iy (I'll be using [ ] as the modulus symbol)

3[(x - 1) + iy]^2 = [(x + 1) + iy]^2

Remember that the modulus squared is equal to the complex no. times its conjugate. (I'll be using [ ] as normal brackets now).

Hence 3[(x - 1) + iy][(x - 1) - iy] = [(x + 1) + iy][(x + 1) - iy]

3[(x - 1)^2 + y^2] = (x+1)^2 + y^2

After performing some rearranging, we arrive at the quadratic

2x^2 - 8x + 2 = 0

Solving this, we get x = 2+sqrt3 and x=2-sqrt3

Since z = x + iy, z= (2+-sqrt3) + 0i

(I'm using [ ] as modulus symbols again)

[z - 2]^2 = [ 2+-sqrt.3 - 2]^2
= [+-sqrt.3]^2
= 3

Hence [z - 2]^2 = 3 for the equation to hold.

 

[dasicmankev]

For the last question, arg[(z-z1)/(z-z2)] = B

This is the same as writing arg(z - z1) - arg(z - z2) = B

The locus z is the arc of a circle with z1 and z2 as the endpoints, the two subtending an angle of B at the circumference. Note that z =/= z1 or z2.

 

[dasicmankev]

For part (iv), you square the equation on both sides and arrive at the same equation as in part (iii). We proved before that the equation only holds if [z - 2]^2 = 3. (using [ ] as modulus symbol)

Hence [z - 2]= sqrt 3

The locus of z is circle with radius sqrt.3, centre (2 + 0i).

For part (v) a,

you let z = x + iy and let w = a + ib

[z + w]^2 + [z - w]^2 = [(x + a) + (y+b)i]^2 + [(x - w) + (y - b)i]^2.
= (x + a)^2 + (y + b)^2 + (x - w)^2 + (y - b)^2.
= 2(x^2) + 2(y^2) + 2(a^2) + 2(b^2)
= 2[(x^2 + y^2) + (a^2 + b^2)]
= 2 [ (mod.(z))^2 + (mod(w))^2 ]

 

[khorne]

Va)

|z|^2 = zz* (z* = conj z)

let z = x+y

|x+y|^2 = (x+y)(x+y)* = (x+y)(x*+y*)

do this for both, expand, collect like terms and you win