# Complex numbers assistance (1 Viewer)

#### epsilon-

##### ?
How can I do this?

|a|^2 + |b|^2 - |a-b|^2 = 2Re(a bconjugate)

Thanks.

#### Sy123

##### This too shall pass
$\bg_white |a-b|^2 = (a-b) \overline{(a-b)} = (a-b)(\bar{a} - \bar{b}) = a\bar{a} - a\bar{b} - b\bar{a} + b\bar{b}$

$\bg_white \Rightarrow \ |a-b|^2 = |a^2| - (a\bar{b} + b\bar{a}) + |b|^2$

$\bg_white |a|^2 + |b|^2 - |a-b|^2 = a\bar{b} + b\bar{a}$

if we let

$\bg_white a=x+iy$

$\bg_white b=u+iv$

It becomes clear that,

$\bg_white a\bar{b} + b\bar{a} = 2(xu+yv) = 2Re(a\bar{b})$

$\bg_white \Rightarrow \ |a|^2 + |b|^2 - |a-b|^2 = 2Re(a\bar{b})$

#### anomalousdecay

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

Hang on. Does the question ask to prove it or to find the locus defined as that?

#### epsilon-

##### ?
$\bg_white |a-b|^2 = (a-b) \overline{(a-b)} = (a-b)(\bar{a} - \bar{b}) = a\bar{a} - a\bar{b} - b\bar{a} + b\bar{b}$

$\bg_white \Rightarrow \ |a-b|^2 = |a^2| - (a\bar{b} + b\bar{a}) + |b|^2$

$\bg_white |a|^2 + |b|^2 - |a-b|^2 = a\bar{b} + b\bar{a}$

if we let

$\bg_white a=x+iy$

$\bg_white b=u+iv$

It becomes clear that,

$\bg_white a\bar{b} + b\bar{a} = 2(xu+yv) = 2Re(a\bar{b})$

$\bg_white \Rightarrow \ |a|^2 + |b|^2 - |a-b|^2 = 2Re(a\bar{b})$
Fanks MOIT! <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.
Fanks MAiTE! <3
Hang on. Does the question ask to prove it or to find the locus defined as that?
No