Complex (1 Viewer)

azureus88 · Mar 25, 2009

If w=cis(2pi/8)=cis(pi/4).

Prove (1-w)(1-w^2)(1-w^3)...(1-w^7)=8

vds700 · Mar 25, 2009

the approach i would use is
$$Let$ z= cis\theta \\1-z=1-cis\theta\\=(1-cos\theta)-isin\theta\\=2sin^{2}\frac{\theta}{2}-2isin\frac{\theta}{2}cos\frac{\theta}{2}\\=2sin\frac{\theta}{2}[sin\frac{\theta}{2}-icos\frac{\theta}{2}]\\=2sin\frac{\theta}{2}[cos(\frac{\pi}{2}-\frac{\theta}{2})-isin(\frac{\pi}{2}-\frac{\theta}{2})]\\=2sin\frac{\theta}{2}[cos(\frac{\theta}{2}-\frac{\pi}{2})+isin(\frac{\theta}{2}-\frac{\pi}{2})]$

now just transform each term into mod-arg form, then its easy to multiply by adding arguments etc

Timothy.Siu · Mar 25, 2009

=-(1-w)(1-w')(1-w^2)(1-w^2')(1-w^3)(1-w^3')
=-(1-(cos pi/4+isin pi/4)(1-(cos pi/4-isin pi/4)(1-(cos pi/2+isin pi/2)(1-(cos pi/2-isin pi/4)(1-(cos 3pi/4+isin 3pi/4)(1-(cos 3pi/4-isin 3pi/4)
=-(1-2cos pi/4+1)(1-2cos pi/2+1)(1-2cos 3pi/4+1)
=-(2-2cos pi/4)(2)(2-2cos 3pi/4)
=-2(4-2)=-4

lol i got owned

Complex (1 Viewer)

azureus88

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vds700

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Timothy.Siu

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