Problem with complex numbers!!!!!help pleaseeeeeeeee!!!!!!!!!?

Question

If for the number zεC is /z-i/=1, prove that 4<=/z+4+2i/<=6


(/..../ is the modulus)

Answer 1

let z = a + bi

|z -i| = 1 ...
|a + (b-1)i| = 1
then a² + (b-1)² = 1. This is a circle with center at (0,1) radius 1.

|z+4+2i| ... this means we need the nearest and farthest distance of our circle from the point (-4, -2) ...
the distance of the center of the circle from the point is...
√{(-4-0)² + (-2-1)²} = 5.

And since the radius is 1 ...
This means the nearest distance is 5 - 1.
The farthest distance is 5 + 1.

§

Answer 2

z=x+iy Iz-iI=1 represents a circle center (0,1) and radius 1
x^2+y^2-2y=0
Ix+4+i(y+2)I ^2 = (x+4)^2+(y+2)^2
Extremes of
(x+4)^2+(y+2)^2 with the condition x^2+y^2-2y=0
F(x,y,k)= (x+4)^2+(y+2)^2+k(x^2+y^2-2y) Lagrangian multipliers
Fx= 2(x+4)+2kx=0
Fy= 2(y+2) +2ky-2k

so x= -4/(1+k)
y=(2k-2)/(1+k)
and
16/(1+k)^2+(2k-2)^2/(1+k)^2 -2(2k-2)/(1+k)=0
16+(2k-2)^2-2(2k-2)*(1+k)=0
From here you get 2 values f k which gives you x and y for extremes
Verify the minimum is 16 and the maximum 36

Answer 3

Sorry, i am afraid of such maths.