# Thread: Complex numbers

1. ## Complex numbers

1. If $\displaystyle z=x+iy$ satisfies $\displaystyle \mid z-4 \mid + \mid z+4 \mid =10$ show that

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

2. Conversely, if x and y satisfy $\displaystyle (x/5)^2+(y/3)^2=1$, then show that $\displaystyle z=x+iy$ satisfies $\displaystyle \mid z-4 \mid + \mid z+4 \mid =10$.

Not a clue how to do this, nothing seems to work.

2. These may help you. You can square both sides.
$\displaystyle \left| {z - 4} \right|^2 = \left( {z - 4} \right)\left( {\overline z - 4} \right) = z\overline z - 4z - 4\overline z + 16$

$\displaystyle \left| {z + 4} \right|^2 = \left( {z + 4} \right)\left( {\overline z + 4} \right) = z\overline z + 4z + 4\overline z + 16$

$\displaystyle \left| {z - 4} \right|^2 + \left| {z + 4} \right|^2 = 2z\overline z + 32$

$\displaystyle \left( {z + 4} \right)\left( {z - 4} \right) = z^2 - 16$

3. Hello, Thomas!

Another approach . . .

If $\displaystyle z\:=\:x+iy$ satisfies $\displaystyle |z-4| + |z + 4|\:=\:10$

show that: .$\displaystyle \left(\frac{x}{5}\right)^2 + \left(\frac{y}{3}\right)^2\:=\:1$
$\displaystyle |z-4|$ is the distance from $\displaystyle z$ to the point (4,0).

$\displaystyle |z+4|$ is the distance from $\displaystyle z$ to the point (-4,0).

If the sum of the distances to two fixed points is a constant,
. . the locus of $\displaystyle z$ is an ellipse, whose equation can be derived
. . with the Distance Formula (and a lot of algebra).

4. Thanks. I can do (1) by taking $\displaystyle \left| z+4 \right|$ to be $\displaystyle \sqrt{(x+4)^2+y^2}$ and the same for the other. However I'm having trouble doing the opposite, starting with the ellipse eqn and finding that $\displaystyle z=x+iy$...