Locus Questions (1 Viewer)

[kevda1st]

i couldnt do these...
3ReZ - 4ImZ =12

2|z|= Z + Z(bar) +4

|Z^2 - [Z(bar)]^2 | < 4

Region common to ZZ(bar) <4 and Z- Z(bar) >2i

thanks in advance

 

[shaon0]

i couldnt do these...
3ReZ - 4ImZ =12

2|z|= Z + Z(bar) +4

|Z^2 - [Z(bar)]^2 | < 4

Region common to ZZ(bar) <4 and Z- Z(bar) >2i

thanks in advance

I'll start you off.
If z=a+ib.

3Re(z) - 4Im(z) = 12
3a - 4b =12

2mod(z) = a+ib a-ib +4
= 2a+4
=2(a+2)

(a^2+b^2)<4 and a>i
Are the common regions.

 

[alakazimmy]

1. 3ReZ - 4ImZ =12
Shaon explained this well.

It is a straight line, with x-intercept = 4, and y-int = -3

2. 2|z|= Z + Z(bar) +4
Let z=a+ib

2 (a^2+b^2)^(1/2) = a + ib + a - ib +4
= 2a + 4 = 2 (a + 2)

2 is a common factor, so we can cancel that out. After that, square both sides

a^2 + b^2 = a^2 + 4a + 4
b^2 = 4(a + 1)

Hence, the locus is a parabola with centre (-1,0) with focus (0,0)

3. |Z^2 - [Z(bar)]^2 | < 4

Again, let z = a+ib

Now (Z^2 - [Z(bar)]^2) is the difference between 2 squares.
So (Z^2 - [Z(bar)]^2) = (z - z(bar)) (z + z(bar))
=(a+ib - (a-ib) ) (a+ib + a - ib)
= (2ib)(2a)
= (4abi)

This number is purely imaginary, so the modulus of it is just the coefficient of i, which is 4ab
Hence, 4ab < 4 ==> ab < 1.

Therefore, the locus the area between two rectangular hyperbola. It includes areas in the 1st and 3rd quadrants (and nothing in the 2nd and 4th)

4. Region common to ZZ(bar) <4 and Z- Z(bar) >2i

first part is a cricle.
The 2nd part is:
2bi > 2i
So b>1

It is the common area of the circle with centre (0,0), radius 2, and the area above the line b=1.