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# Complex Function help

Thu 23 Aug, 2012 08:31 pm
For the function g(w)=√2/2−√2/2e^(iw), show that;

l G(w) l^2 + l G(w+pi ) l^2 = 2

Can anyone help me out with this im really confused and any help would be great.
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uvosky

1
Fri 24 Aug, 2012 04:48 am
@John-kash,
You first used g(w) and then used G(w) , do you want to mean the same
function by both symbol ? or is G(w) the generating function of g(w) ?
or something else ?
John-kash

1
Fri 24 Aug, 2012 07:30 am
@uvosky,
sorry, they are both meant to be upper case 'G' to mean the one function.
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John-kash

1
Fri 24 Aug, 2012 09:43 pm
@John-kash,
Any help anyone?
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uvosky

1
Mon 27 Aug, 2012 05:29 am
@John-kash,
Here's the story , g(w)= 1/sqrt2 - 1/(sqrt2) e^iw = 1/sqrt2 - e^-iw/sqrt2
= 1/sqrt2 - cosw/sqrt2 + isinw/sqrt2
so , lg(w)l^2 = (1/2) { (1-cosw)^2 + (sinw)^2 }
since e^-i(w+pi) = -cosw + isinw we have ,
g(w+pi) = 1/sqrt2 + cosw/sqrt2 - isinw/sqrt2
so , lg(w+pi)l^2 = (1/2) { (1+cosw)^2 + (sinw)^2 } whence we get
lg(w)l^2 + lg(w+pi)l^2 = (1/2) { 2 + 2 (cosw)^2 + 2(sinw)^2 }
= (1/2) (2+2) = 2
(by sqrt2 I mean square root of 2)
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