ultra hard complex no. (1 Viewer)

[namburger]

http://i237.photobucket.com/albums/ff3/Namburger/Complexnoq2.jpg

umm yeh, i can do i) and ii). No problem

I can do iii) and iv) by otherwise method which takes a million years, can someone help me by using the "hence"


 

[Slidey]

Underlined = conjugate.

1) z1 = a+ib, so
|z1| = sqrt(a^2+b^2)
RHS =|z1|^2 = a^2+b^2
LHS = z1*z1 = (a+ib)(a-ib) = a^2+b^2 #

2) it's fucking obvious. I mean the conjugate of a complex number is itself still a complex number (and therefore has a conjugate), making it a trivial rewording of the rule. I hate that kind of question. If they want to use it as a lead in, they should say "by using rule a or otherwise, prove:". Use the Cartesian form of a complex number, as above, to 'prove' it, I suppose.

3) From 1), and with z1=z, z2=w
RHS = |z+w|^2 = (z+w)(z+w) = (z+w)(z+w)
= zz + zw+wz + ww
= |z|^2 + (zw+zw) + |w|^2 = LHS
(because (zw+zw)=Re(zw), which you can confirm by expanding with z=a+ib, w=c+id)

4) |z|^2 >= (Re(z))^2 because a^2+b^2 >= a^2
Similarly, |zw|^2 >= (Re(zw))^2
.'. |z+w|^2 = |z|^2 + |w|^2 + 2Re(zw) <= |z|^2 + |w|^2 + 2|zw|
Since |zw|=|zw|=|z||w|,
|z|^2 + |w|^2 + 2|zw| = (|z|+|w|)^2
,', |z+w|^2 <= (|z|+|w|)^2
So |z+w| <= |z|+|w|

 

[Poad]

Underlined = complement.

I suppose you mean 'conjugate'? >_>

 

[Slidey]

God damn it, I've got statistics on the brain. Yeah, conjugate.

>_>

And don't take that tone with me.

 

[Poad]

Oops, guess that got taken the wrong way, I meant it in no patronising tone whatsoever (which is how I assume you took it).

And besides, I tend to overuse that smiley alot. >_> ...Damnit

 

[Slidey]

Haha I'm just being silly. I know you meant no disrespect. I do think that smiley says "I'm frowning at your stupidity", however.

 

[undalay]

I do think that smiley says "I'm frowning at your stupidity", however.

I think this one is more so:

-______-"""

 

[tommykins]

回复: Re: ultra hard complex no.

._.

 

[Slidey]

Re: 回复: Re: ultra hard complex no.

_-_