E epsilon- ? Joined Oct 11, 2012 Messages 26 Gender Male HSC 2013 Apr 10, 2013 #1 How can I do this? |a|^2 + |b|^2 - |a-b|^2 = 2Re(a bconjugate) Thanks.
Sy123 This too shall pass Joined Nov 6, 2011 Messages 3,734 Gender Male HSC 2013 Apr 10, 2013 #2 if we let It becomes clear that,
anomalousdecay Premium Member Joined Jan 26, 2013 Messages 5,778 Gender Male HSC 2013 Apr 10, 2013 #3 Let: a=x + iy, b= x - iy |a|^2 = x^2 + y^2 |b|^2 = x^2 + y^2 a-b = 2yi |a-b| = sqrt(4y^2) = 2y 2x^2 + 2y^2 + 2y = 2Re(x^2 - y^2) = x^2 -y^2 3y^2 + x^2 + 2y = 0 The locus of a is: 3y^2 + x^2 + 2y = 0, which represents an ellipse.
Let: a=x + iy, b= x - iy |a|^2 = x^2 + y^2 |b|^2 = x^2 + y^2 a-b = 2yi |a-b| = sqrt(4y^2) = 2y 2x^2 + 2y^2 + 2y = 2Re(x^2 - y^2) = x^2 -y^2 3y^2 + x^2 + 2y = 0 The locus of a is: 3y^2 + x^2 + 2y = 0, which represents an ellipse.
anomalousdecay Premium Member Joined Jan 26, 2013 Messages 5,778 Gender Male HSC 2013 Apr 10, 2013 #4 Hang on. Does the question ask to prove it or to find the locus defined as that?
E epsilon- ? Joined Oct 11, 2012 Messages 26 Gender Male HSC 2013 Apr 10, 2013 #5 Sy123 said: if we let It becomes clear that, Click to expand... Fanks MOIT! <3 anomalousdecay said: Let: a=x + iy, b= x - iy |a|^2 = x^2 + y^2 |b|^2 = x^2 + y^2 a-b = 2yi |a-b| = sqrt(4y^2) = 2y 2x^2 + 2y^2 + 2y = 2Re(x^2 - y^2) = x^2 -y^2 3y^2 + x^2 + 2y = 0 The locus of a is: 3y^2 + x^2 + 2y = 0, which represents an ellipse. Click to expand... Fanks MAiTE! <3 anomalousdecay said: Hang on. Does the question ask to prove it or to find the locus defined as that? Click to expand... No
Sy123 said: if we let It becomes clear that, Click to expand... Fanks MOIT! <3 anomalousdecay said: Let: a=x + iy, b= x - iy |a|^2 = x^2 + y^2 |b|^2 = x^2 + y^2 a-b = 2yi |a-b| = sqrt(4y^2) = 2y 2x^2 + 2y^2 + 2y = 2Re(x^2 - y^2) = x^2 -y^2 3y^2 + x^2 + 2y = 0 The locus of a is: 3y^2 + x^2 + 2y = 0, which represents an ellipse. Click to expand... Fanks MAiTE! <3 anomalousdecay said: Hang on. Does the question ask to prove it or to find the locus defined as that? Click to expand... No
anomalousdecay Premium Member Joined Jan 26, 2013 Messages 5,778 Gender Male HSC 2013 Apr 14, 2013 #6 So are we both correct in that sense? Wow I have never come across general questions like that before.
So are we both correct in that sense? Wow I have never come across general questions like that before.