I try to solve the following task:
Show with Euler's formula (from complex analysis) that for z≠2kπ where k∈Z the following is true: n∑v=0cos(vz)=12+12sin(nz+z2)sinz2
My attempt:
2n∑v=0cos(vz)=1+sin(nz+z2)sinz2
n∑v=02cos(vz)−n∑v=01n=sin(nz+z2)sinz2
n∑v=02cos(vz)sinz2−sinz2n=sin(nz+z2)
Now I tried using the sinus identity sin(x±y)=sinxcosy±cosxsiny
My second thought to write cosinus as a series doesn't look like it will help either:
2n∑v=0∞∑m=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)=ℜeivz
n∑v=0cos(vz)=ℜn∑v=0eivz=ℜn∑v=0(eiz)v
Sum of exponential:
=ℜ1−(eiz)n+11−eiz
Can you take it from here?
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