Complex numbers

**Thomas154321** · October 28th 2007, 03:14 PM

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

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

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

**Plato** · October 28th 2007, 04:00 PM

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

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

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

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

**Soroban** · October 28th 2007, 07:51 PM

Hello, Thomas!

Another approach . . .

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

show that: . $\left(\frac{x}{5}\right)^2 + \left(\frac{y}{3}\right)^2\:=\:1$

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

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

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

**Thomas154321** · October 29th 2007, 12:20 AM

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

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