Question

Having difficulty with the following:

http://i.imgur.com/xK02a.png

Thanks in advance :)

Answer 1

at which step are u stuck?
u know the value of e^(i*pi)

Answer 2

Help you in a minute

Answer 3

Conjugate of Gw\[\frac{\sqrt2}{2}-\frac{\sqrt2}{2}cosw+i\frac{\sqrt2}{2}sinw\], as the imaginary part is reversed in a conjugate
Hw is\[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}cosw+i\frac{\sqrt2}{2}sinw\]

Answer 4

\[e^{ix}=cosx+isinx\]\[cosx=\Re (e^{ix})\] Squiggly r=real part of
\[cos(a+b)=\Re (e^{i(a+b)})=\Re (e^{ia}e^{ib})\]

Answer 5

\[=\Re((cosa+isina)(cosb+isinb))=cosacosb-sinasinb\]
So \[\cos(w+\pi)=cosw (-1)-sinw(0)=-cosw\]
Likewise
\[\sin(a+b)= Im(e^{i(a+b)})\]

Answer 6

\[\sin(w+\pi)=\sin(w)\cos(\pi)+\sin(\pi)\cos(w)=-\sin(w)\]

Answer 7

So H(w+pi):\[\frac{\sqrt2}{2}-\frac{\sqrt2}{2}cosw-i\frac{\sqrt2}{2}sinw\], conjugate of G(w+pi):\[\frac{\sqrt2}{2}+\frac{\sqrt2}{2}cosw-i\frac{\sqrt2}{2}sinw\]. Now multiply all the appropriate things together (use wolfram, it'll be messy)

Answer 8

\[e^{ix}=cosx+isinx\]is key here

Answer 9

Ask anything if you don't get it.

Answer 10

Thanks so much henpen, really appreciate that.