Question

If someone could provide assistance I would greatly appreciate this.

For the function G(w) = (sqrt2/2) - (sqrt2/2 e^iw)

show that: Re(G(w)) = sqrt2 sin^2 (w/2) and Im(G(w)) = -1/sqrt2 sin w

Answer 1

\[G(w) = \frac{\sqrt2}2 -\frac{\sqrt2}{2 e^{iw}}\]\[=\frac{\sqrt 2}2(1-e^{-iw})\]\[=\frac{\sqrt 2}2\left(1-\left(\cos(-w)-i\sin(-w\right)\right))\]\[=\frac{\sqrt 2}2\left(1+\cos(w)-i\sin(w\right))\]

Answer 2

Thank you kindly for your help, however, I mistakenly entered the question wrong.

It should read:

\[G(w) = \frac{ \sqrt{2} }{ 2 } - \frac{ \sqrt{2} }{ 2 } e ^{iw}\]

then show that: \\[\Re(G(w) = \sqrt{2}\sin ^{2}\frac{ w }{ 2 }\]

and \[\Im(G(w)) =-\frac{ 1 }{ \sqrt{2} }\sin w \]

Answer 3

What math are you in?

Answer 4

Calc. IIA

Do you have any idea for the above?

Answer 5

I took calc 2 last sem, what chapter is it? or what is the section called and i will look at my notes

Answer 6

Oh, it was brought up in class as a problem question. Unfortunately, cant find any reference.

Answer 7

well whats the name of the chapter you are learning

Answer 8

\[G(w) = \frac{\sqrt2}2 -\frac{\sqrt2}{2 }e^{iw}\]\[e^{iw}=\cos w+i \sin w\]\[G(w) = \frac{\sqrt2}2 (1-e^{iw})=\frac{\sqrt2}2 (1-\cos w+i \sin w)=\frac{\sqrt2}2 (2\sin^2 \frac{w}{2}+i \sin w)\]does this help?

Answer 9

oops...sorry
\[G(w) = \frac{\sqrt2}2 (1-e^{iw})=\frac{\sqrt2}2 (1-\cos w-i \sin w)=\frac{\sqrt2}2 (2\sin^2 \frac{w}{2}-i \sin w)\]

Answer 10

fantastic, thank you so much :)