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For the function $$G(w) = \frac{\sqrt2}{2}-\frac{\sqrt2}{2}e^{iw},$$ show that $$G(w) = -\sqrt2ie^{iw/2} \sin(w/2).$$

Hey everyone, I'm very new to this kind of maths and would really appreciate any help. Hopefully I can get an idea from this and apply it to other similar questions. Thank you.

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  • $\begingroup$ Are you familiar with $e^{i\theta}=\cos\theta+i\sin\theta$? $\endgroup$ – Gerry Myerson Aug 20 '12 at 7:20
  • $\begingroup$ @GerryMyerson Will it work? The exp(w) -> sin(w/2) via Euler's doesn't halve the angle while introduces a cosine. Am I missing something? $\endgroup$ – Frenzy Li Aug 20 '12 at 7:25
  • $\begingroup$ @Frenz, I just want to know whether OP is familiar with that formula. If not, then the answer posted by Euler is unlikely to be helpful, and we know where an answer has to start. If OP is familiar with that formula, then we can show how to get from there to Euler's answer. $\endgroup$ – Gerry Myerson Aug 20 '12 at 7:36
  • $\begingroup$ @GerryMyerson Oh, right. The worded familiar and question mark. :) $\endgroup$ – Frenzy Li Aug 20 '12 at 7:42
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Use the definition for the complex sine: $$ \sin(z)=\frac{ e^{iz}-e^{-iz} } {2i} $$ Thus, $$-\sqrt{2}ie^{i\frac{w}{2}}\sin\frac{w}{2} =-\sqrt{2}ie^{i\frac{w}{2}}(\frac{1}{2i}(e^{i\frac{w}{2}} - e^{-i\frac{w}{2}})) $$

Now simplify to get your result.

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  • $\begingroup$ @FrenzY DT. Thanks, I was told that if i use the complex sine for this, it = -isinh(i*w/2). But im not really sure about that and where to go from here. Sorry if this is obvious to you. $\endgroup$ – john kash Aug 21 '12 at 1:29

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