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Complex number identities

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Let z, w E C

a) Prove the following identities:

i) |z+w|^2 = |z|^2 + 2Re(zw) + |w|^2
ii) |z - w|^2 = |z|^2 - 2Re(zw) + |w|^2

b) Deduce that |z+w|^2 + |z-w|^2 = 2(|z|^2 + |w|^2).

c) Use (a)(i) to prove that |z+w| < |z-| + |w| and give necessary conditions for equality to hold.

d) Prove that [|z| - |w|] < |z-w|.

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Solution Summary

Some complex number identities are proven in the solution.

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Let z = x1+i.y1, w = x2 + i.y2

a.i.
Left Hand Side (L.H.S.):
|z+w|^2 = (x1+x2)^2 + (y1+y2)^2 = x1^2 + x2^2 + y1^2 +y2^2 + 2x1.x2 + 2y1.y2
= x1^2 + y1^2 + x2^2 + y2^2 + 2(x1.x2+y1.y2)
= |z|^2 + |w|^2 + 2Re(z.w_bar) = R.H.S.

[Because, z.w_bar = (x1+i.y1).(x2-i.y2) = (x1.x2 + y1.y2) + i.(x2.y1 - x1.y2)]

a.ii.
L.H.S.:
|z-w|^2 = (x1-x2)^2 + (y1-y2)^ = x1^2 + x2^2 + y1^2 +y2^2 - 2x1.x2 - 2y1.y2
= x1^2 + y1^2 + x2^2 + y2^2 - 2(x1.x2+y1.y2)
= |z|^2 + |w|^2 - 2Re(z.w_bar) = R.H.S.

b.
L.H.S.
|z+w|^2 + |z-w|^2 = ...

Solution provided by:
Amrit Lal Ahuja, PhD (IP)
Education
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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