# (fp2) simplifying equations with modulii (modulus'?)Watch

#1
|x + iy - 2i| = 2|x + iy + i|
|x+i(y-2)| = 2|x+i(y+1)|
x^2 + (i^2)(y-2)^2 = 2(x^2 + (i^2)(y+1)^2)
x^2 - (y-2)^2 = 2(x^2 - (y+1)^2)
x^2 - (y-2)^2 = 2x^2 - 2(y+1)^2

anyone know where i went wrong with my simplifying?
0
7 years ago
#2
(Original post by anus eater)
|x + iy - 2i| = 2|x + iy + i|
|x+i(y-2)| = 2|x+i(y+1)|
x^2 + (i^2)(y-2)^2 = 2(x^2 + (i^2)(y+1)^2)
x^2 - (y-2)^2 = 2(x^2 - (y+1)^2)
x^2 - (y-2)^2 = 2x^2 - 2(y+1)^2

anyone know where i went wrong with my simplifying?
The square of the modulus of a+ib is a^2+b^2; you don't do anything with the "i". You're simply using pythagoras.

You also forgot to square the "2" outside the modulus sign.
#3
(Original post by ghostwalker)
The square of the modulus of a+ib is a^2+b^2; you don't do anything with the "i". You're simply using pythagoras.

You also forgot to square the "2" outside the modulus sign.
ah, makes perfect sense.

and yeah i only just noticed that too lol, thanks!
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