"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is. ” J. von Neumann
nice complex exercise asked by Dan Stirua in many math groups
Exercise:
Show that , |n∑i=1(Re(zi)−Im(zi))×n∑i=1(Re(zi)+Im(zi))|(n∑i=1|zi|)2≤2 , {zi}i≥1∈Cn
Solution : Let z=a+ib⇔ˉz=a−ib
So z+ˉz=(a+ib)+(a−ib)=2a=2Re(z) and z−ˉz=a+ib−a+ib=2ib=2iIm(z)
Hence Re(z)=z+ˉz2&Im(z)=z−ˉz2i
So Re(z)+Im(z)=z+ˉz2+−iz+iˉz2=1−i2z+1+i2ˉz and
Re(z)−Im(z)=z+ˉz2−(−iz+iˉz2)=1+i2z+1−i2ˉz
Let A=n∑k=1(Re(zk)−Im(zk))&B=n∑k=1(Re(zk)+Im(zk))
So A=n∑k=1((1+i2)zk+(1−i2)ˉzk)=(1+i2)n∑k=1zk+(1−i2)n∑k=1ˉzk and
B=n∑k=1((1−i2)zk+(1+i2)ˉzk)=(1−i2)n∑k=1zk+(1+i2)n∑k=1ˉzk
So |A|=|(1+i2)n∑k=1zk+(1−i2)n∑k=1ˉzk|≤|1+i2||n∑k=1zk|+|1−i2||n∑k=1ˉzk|
But |1+i2|=|1−i2|=√22=1√2 and |zk|=|ˉzk| so A≤2√2n∑k=1|zk|
Similarly , |B|≤2√2n∑k=1|zk| Thus |AB|≤|A||B|=2(n∑k=1|zk|)2
Therefore |AB|(n∑k=1|zk|)2≤2 Q.E.D
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