Complex Number Locus Help~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ (1 Viewer)

[mike7955]

How do you sketch the followings?

|z-(3+2i)| + |z-(-3+3i)| = 7

Im((z-1+4i)^2)=1

arg[(z-z0)/(z1-z0)]=0
where z0 = -4-3i and z1 = -2+4i

|Re(z)|<4 and |Im(z)|<4

|Re(z)|>1 and |Im(z)|>1

 

[braintic]

1. An ellipse with foci at (3+2i) and (-3+2i) and a major axis of length 7 units. (using sum of focal lengths on an ellipse = 2a)

2. Im [(x-1) + i(y+4) ]^2 = 1
Im [(x-1)^2 +2i(x-1)(y+4) - (y+4)^2] = 1
2(x-1)(y+4) = 1
hyperbola, asymptotes x=1, y=-4

3. Using arg(z1/z2) = arg(z1) - arg(z2), equation becomes arg(z-z0) = arg(z1-z0)=arg(2+7i) = ? (calculator work)
so ray, starting from z0, making angle of arctan(7/2) with +ve x-axis

4. |x|<4 and |y| < 4
So, a filled in square bounded by the dashed lines x=4, x=-4, y=4, y=-4

5. Same idea as 4, but (extending the sides of the square) the regions adjacent to the 4 corners of the square are shaded.

 

[anomalousdecay]

1) Let P be the locus, a be the major axis, b be the major axis and S,S' be the foci. PS + PS' = 2a = 7 (Given Data) (Definition of an ellipse) a=3.5, therefore a^2 = 12.25
Now, SB = S'B = 3.5 (Using pythagoras' theorem) b= sqrt(3.5^2 - 3^2) = 3.25 b^2 = 10.5625
The Locus of P is the hyperbola (x/ 12.25)^2 + (y/10.5625)^2 = 1.

The rest are correct by braintic.