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Complex numbers help (1 Viewer)

beolun

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Prove cisθ/(1-cisθ)=i(cis(θ/2))/(2sin(θ/2)).
 

hit patel

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Well the way I would do it is instead of cis theta write cos theta + isin theta and then apply the double angle formula for each term i,e, isin theta becomes 2i sin theta/2 cos theta/2. Then simplifying it. Sorry i wasnt able to show it. This is how used used the trigonometric polar fractions in Demoivre theorem section but this doesnt consist of a power.
 

beolun

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I tried that and didn't get the required result :cry:, did you actually get to the answer?
 

hit patel

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dont have a pen or paper on hand currently therefore I couldnt do it as I mentioned. I might try when i get home if no one responds.
 

HeroicPandas

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Assuming you know all your trig identities....

Let z = cis(theta)



note: z . z(bar) = |z|^2



I didnt use the double angle (or half angle) of sine on the numerator because the way i was gonna use it may confuse u
 

hit patel

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Damn....... Heroic u genius. Why didnt I figure that out? HOw did you know what to do?
 

HeroicPandas

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I am no genius, look at the denominator of the answer, there is no imaginary so you must 'realise' the denominator (like rationalising - by timing top and bot by conj)

And the rest is trig identities (double/half angles) and trying to make LHS look like the RHS, by taking out the half, by factorising 'i', blah blah

If you know your basics, then everything will be fine!
 

hit patel

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HMMM..... Yep that is true>>>> Genius as in like nice work. What mark you aiming for in MX2?
 

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