Saturday, 11 July 2015

complex analysis - Applying Euler's formula



I try to solve the following task:




Show with Euler's formula (from complex analysis) that for z2kπ where kZ the following is true: nv=0cos(vz)=12+12sin(nz+z2)sinz2



My attempt:
2nv=0cos(vz)=1+sin(nz+z2)sinz2



nv=02cos(vz)nv=01n=sin(nz+z2)sinz2



nv=02cos(vz)sinz2sinz2n=sin(nz+z2)



Now I tried using the sinus identity sin(x±y)=sinxcosy±cosxsiny

on the term cos(vz)sinz2 but It doesn't seem to make it easier nor help.




My second thought to write cosinus as a series doesn't look like it will help either:
2nv=0m=0(1)m(vz)2m(2m)!=1+sin(nz+z2)sinz2



I also don't see where I can apply the Euler formula.



Thanks for help


Answer



cos(vz)=eivz




nv=0cos(vz)=nv=0eivz=nv=0(eiz)v



Sum of exponential:



=1(eiz)n+11eiz



Can you take it from here?


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