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Complex Q. (1 Viewer)

KFunk

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There may well be better ways to do this but this seems to work.

If Z = cos@ + isin@ then 1/(1+z) is equal to,

1/[ (1 + cos@) + isin@) * [(1+cos@)-isin@]/[(1+cos@)-isin@]

(to make the denominator real)

= [cos@ +1 - isin@]/[2cos@ + 2]

= 1/2{1 - i(sin@)/(cos@ + 1)}


now consider (sin@)/(cos@ + 1) , letting A = @/2

(sin@)/(cos@ + 1) = 2sinAcosA/(cos<sup>2</sup>A - sin<sup>2</sup>A +1)

= 2sinAcosA/2cos<sup>2</sup>A
= tanA
= tan(@/2)

&there4; 1/(1+z) = 1/2[ 1 - itan(@/2)]
 

gman03

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z=ei@


1/(1+z)

= 1 / (1+ei@)

= e-i@/2 / (e-i@/2+ei@/2)

= [ cos @/2 - i sin @ /2 ] / (2 cos @/2)

= 1/2 - i/2 tan @/2
 

brett86

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Trev said:
we never really did anything with Euler's theorem.
the euler notation shouldnt stop u from using gman03's method, it can easily be re-written in the hsc's cis form

e.g.

 

brett86

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Stefano said:
brett86,

How did you know you should multipy by cis(-@/2) ??
multiplying by cis(-@/2) was gman03's idea, i just rewrote his solution using cis instead of e^(i@)

however, gman03 probably realised that by multiplying the bottom of the fraction by cis(-@/2) he could make it real and simplify the problem

e.g. cis @ + cis (-@) = 2 cos @

here's another question that uses this technique, give it a try:

 

brett86

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hmm, not sure

id use the cis notation just to be safe

the e^(i@) notation does make things alot neater though, its certainly alot better than cis.

in fact i havent seen cis used anywhere else besides the hsc

based on how pointless the cis notation is, it was probably created by the board of studies...
 

justchillin

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Apparently some pple dont like CIS notation either...to be safe id recommend writing out (cos + isin)...keep the marker onside.
 

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