# Complex numbers

• October 28th 2007, 03:14 PM
Thomas154321
Complex numbers
1. If $z=x+iy$ satisfies $\mid z-4 \mid + \mid z+4 \mid =10$ show that

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

2. 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.:o
• October 28th 2007, 04:00 PM
Plato
$\left| {z - 4} \right|^2 = \left( {z - 4} \right)\left( {\overline z - 4} \right) = z\overline z - 4z - 4\overline z + 16$

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

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

$\left( {z + 4} \right)\left( {z - 4} \right) = z^2 - 16$
• October 28th 2007, 07:51 PM
Soroban
Hello, Thomas!

Another approach . . .

Quote:

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).

• October 29th 2007, 12:20 AM
Thomas154321
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 $z=x+iy$...:(