complex numbers help?

Question

if z=x+iy satisfies |z-4|+|z+4|=10
then show that (x/5)^2 + (y/3)^2 = 1

Answer 1

Subtract |z+4| from both sides and square to get:

|z-4|^2 = 100 - 20 |z+4| + |z+4|^2

But |z-4|^2 = (x-4)^2 + y^2 = x^2-8x + 16 + y^2
|z+4|^2 = (x+4)^2 + y^2 = x^2 + 8x + 16 + y^2

Re-arranging, we get:

20 |z+4| = 100 - 16x

Dividing by 4, we get:

5 |z+4| = 25 - 4x

Squaring again, we get:

25 |z+4|^2 = 625 - 200x + 16x^2

|z+4|^2 = x^2 + 8x + 16 + y^2

So:

25 | z+4|^2 = 25x^2 + 200x + 400 + 25y^2

Combining terms, we get:

9x^2 + 25y^2 = 225

Since 225=9 * 25, when we divide by 225 we get:

(x/5)^2 + (y/3)^2 = 1

Answer 2

This equation says that, on the complex plane, the distance from z to the complexes (4,0) and(-4,0) is 10 So, z is on the ellipse whose foci are (4,0) and(-4,0) and whose semi major axis is 10/2 = 5.

To find the semi minor axis, we have to compute the intersection of the ellipse with the y-axis. If x = 0, then we get the complex iy, and its distance to (-4,0) and (4,0) is sqrt(y^2+ 16). So, 2 sqrt(y^2 + 16) = 10 => y^2 + 16 = 25 => y =3

Then, the major and minor semi axis of the ellipse are 5 and 3. According to the equation of an ellipse, it follows that
(x/5) + (y/3)^2 = 1

Answer 3

z lies on an ellipse, with foci at z = - 4 and z = + 4.

This means that if the semi-axes of the ellipse are a, b and the eccentricity is e : ae = 4.

Also the value 10 tells you that, at the end of the major axis,
a + ae + (a - ae) = 10.

We conclude that a = 5:.
Also b(^2) = a(^2)[1 -e(^2)], which gives b = 3.

The equation of the ellipse is (x/5)^2 + (y/3)^2 = 1.