complex no. question (1 Viewer)

[velox]

1. Prove that
____ _
w.w = w.w (conjugate of (w+w) = w . conjugate of w)


for all complex number w


2. Prove that

____ _
w+w = w+ w (conjugate of (w+w) = w + conjugate of w)

for


for all complex number w

Thanks guys

EDIT: Clarified question

 

[Slidey]

Let w_bar=conjugate of w.

Actually, fuck that. Let me write it out by hand. That's what a tablet PC is for, right?

 

[Slidey]

The last proof contains a typographical error. At the bottom, the RHS should read "2x-2iy-LHS", and before that, it should say "RHS=...=LHS", not "RHS=...=RHS".

Also, if this is not the question you are asking, the general method remains the same.

It probably would have been better to prove them for w and z, and then you'd simply need to replace z with w to see it works for w.w and w+w as well. That is easy enough, though - try it.

 

[velox]

hey slide, thanks but i think i screwed up writing the question. I added a word equation for the first one. The second one is supposed to be the conjugate of (w+w) = the conjugate of w, +w.

 

[Xayma]

2. Prove that

____..........._
w+w = w+ w for


for all complex number w

Thanks guys

EDIT: Clarified question

Let w=i

LHS=
____...__
w+w=2i
=-2i

Now
RHS=
_
w+w=-i+i
=0

LHS≠RHS

∴

____..........._
w+w = w+ w is not true for all complex w (or did I miss something, been awhile since Ive seen complex)

 

[velox]

Hmm my friend said he proved it true for all complex w. Damn

 

[:: ck ::]

conj (w+w) = conj(w) + w

Let w = rcis@
LHS = conj (rcis@ + rcis@)
=r * conj(cis@ + cis@)
= r * conj(2cos@ + 2isin@)
= 2rcis(-@)

RHS = rcis(-@) + rcis@
= r(cis (-@) + cis(@) )
= r(cos@ - isin@ + cos@ + isin@)
= 2rcos@

LHS doesn equal RHS

fix up the question ??

 

[Xayma]

Hmm my friend said he proved it true for all complex w. Damn

It proves true for all real. Thats it.

 

[:: ck ::]

agreed with above

 

[Slidey]

1. Prove that
____ _
w.w = w.w (conjugate of (w+w) = w . conjugate of w)


for all complex number w


2. Prove that

____ _
w+w = w+ w (conjugate of (w+w) = w + conjugate of w)

for


for all complex number w

Thanks guys

EDIT: Clarified question

Let conjugate w=h:
Want: (w+w)_conjugate=w+h
LHS=2h
RHS=w+h
Let w=x+iy
LHS=2x-2iy
RHS=2x
Therefore the only time this could be true is if y, or Re(w), is 0. Id est: if the number is purely real. Xayma and Ryan are correct, and I for one am GLAD you couldn't prove it to be true for complex vlaues of w.

And the first one is, I presume meant to read:
(w.w)_conjugate=w.h
LHS=(w^2)_conjugate=h^2
RHS=wh
LHS=x^2-y^2-2xyi
RHS=x^2-y^2
So we have that this is true for all values of w where Im(w)=0 OR Re(w)=0. Id est: all imaginary and real values of w, but no complex values.
Exempli gratia: 2, 2i, -2, -2i, 5, 6, -i, i, 33 billion * i, but NOT 4+3i.