Expert: Ahmed Salami - 12/27/2005

Question
Dear Mr Salami,

I am a little suck by this question:

Derive the loci (in terms of x and y, the real and imaginary parts respectively of z) defined by:

[z-3]+[z+3]=10

where z is a complex number and the '[' and ']' are magnitude symbols.

I think it describes an ellipse but I am unable to get the answer out.

Thank you very much for your time any help would be apprecited

Answer
Hi Andrew,
Sorry for the time taken.
This happens to be such a tedious one so you'll have to
be patient and careful to see it out.
For a complex number, z = x + iy
where x and y are the real and imaginary parts respectively
The modulus, |z| = sqrt(x^2 + y^2)
Therefore,
|z-3| = sqrt[(x-3)^2 + y^2]
|z+3| = sqrt[(x+3)^2 + y^2]
And so for the equation
|z-3| + |z+3| = 10
sqrt[(x-3)^2 + y^2] + sqrt[(x+3)^2 + y^2] = 10
This is where the long simplification begins, i would skip some of them hoping you'll get by.
squaring both sides, we get
[(x-3)^2 + y^2] + [(x+3)^2 + y^2]
+ 2sqrt[((x-3)^2 + y^2)((x+3)^2 + y^2)] = 100
The term ((x-3)^2 + y^2)((x+3)^2 + y^2)
= (x-3)^2.(x+3)^2 + (x-3)^2.y^2 + (x+3)^2.y^2 + y^4
= (x^2 - 9)^2 + (x^2 - 6x + 9)y^2 + (x^2 + 6x + 9)y^2 + y^4
= (x^4 - 18x^2 + 81) + (2x^2 + 18)y^2 + y^4
= (x^4 - 18x^2 + 81) + (2x^2y^2 + 18y^2) + y^4
= x^4 - 18x^2 + 81 + 2x^2y^2 + 18y^2 + y^4
Going back to our equation, substitution gives
[(x-3)^2 + y^2] + [(x+3)^2 + y^2]
+ 2sqrt(x^4 - 18x^2 + 81 + 2(xy)^2 + 18y^2 + y^4) = 100

(x^2 - 6x + 9 + y^2) + (x^2 + 6x + 9 + y^2)
+ 2sqrt(x^4 - 18x^2 + 81 + 2(xy)^2 + 18y^2 + y^4) = 100

2x^2 + 2y^2 + 18 +
2sqrt(x^4 - 18x^2 + 81 + 2(xy)^2 + 18y^2 + y^4) = 100

2sqrt(x^4 - 18x^2 + 81 + 2(xy)^2 + 18y^2 + y^4)
= 82 - 2x^2 - 2y^2
dividing both sides by 2
sqrt(x^4 - 18x^2 + 81 + 2(xy)^2 + 18y^2 + y^4)
= 41 - x^2 - y^2
squaring both sides,
x^4 - 18x^2 + 81 + 2(xy)^2 + 18y^2 + y^4
= 1681 + x^4 + y^4 - 82x^2 - 82y^2 + 2(xy)^2
Therefore,
64x^2 + 64y^2 = 1600
x^2 + y^2 = 25
Which is the equation of a circle with centre at the origin and a radius of 5 units.
I hope i have helped, you can always get back to me.
Regards.