# Maths helP!

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#1
|z| = |z+2|

show that the real component of z = -1 (done that)

z also satisfies |z| = 2, sketch two loci to find the two possible values of Z

I found them to be z = -1 +/- rt2 * i.

but I don't think that's right, could someone see what they get?
0
16 years ago
#2
(Original post by fishpaste)
|z| = |z+2|

show that the real component of z = -1 (done that)

z also satisfies |z| = 2, sketch two loci to find the two possible values of Z

I found them to be z = -1 +/- rt2 * i.

but I don't think that's right, could someone see what they get?
Hmm that doesn't look right,

say z = x+iy.

Then (x+iy)^2 = ((x+2)+iy)^2.

So x^2-y^2 +2xyi = x^2+4x+4+2(x+2)yi - y^2.

Equating the real and complex parts:

4x = -4. So x=-1.

Given that |z| = 2.

4 = x^2+y^2. So y = +/- root 3.
0
#3
(Original post by theone)
Hmm that doesn't look right,

say z = x+iy.

Then (x+iy)^2 = ((x+2)+iy)^2.

So x^2-y^2 +2xyi = x^2+4x+4+2(x+2)yi - y^2.

Equating the real and complex parts:

4x = -4. So x=-1.

Given that |z| = 2.

4 = x^2+y^2. So y = +/- root 3.
Hm, bad algebra on my part.

Thank you so much 0
16 years ago
#4
(Original post by fishpaste)
|z| = |z+2|

show that the real component of z = -1 (done that)

z also satisfies |z| = 2, sketch two loci to find the two possible values of Z

I found them to be z = -1 +/- rt2 * i.

but I don't think that's right, could someone see what they get?
z = x + iy
=> x^2 + y^2 = (x+2)^2 + y^2
=>...=> x = -1

Loci of this is the line real part = -1 on argand

Loci of lzl = 2 is a crcle centre origin radius 2

loci intersect when x=-1 hence y = +sqrt3

z = -1 + i.sqrt3
0
16 years ago
#5
what is this ? p what?
0
16 years ago
#6
(Original post by kikzen)
what is this ? p what?
P6 for edexcel, although it only uses p4 principles.
0
16 years ago
#7
(Original post by theone)
P6 for edexcel, although it only uses p4 principles.
is there any p6 this month ???
0
#8
P4 Ocr.
0
16 years ago
#9
(Original post by theone)
P6 for edexcel, although it only uses p4 principles.
this is all p4 stuff.... u don't need p6 to know this...
0
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