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nice complex exercise asked by Dan Stirua in many math groups



Exercise:

Show that , |ni=1(Re(zi)Im(zi))×ni=1(Re(zi)+Im(zi))|(ni=1|zi|)22   ,   {zi}i1Cn

Solution : Let z=a+ibˉz=aib

So z+ˉz=(a+ib)+(aib)=2a=2Re(z) and zˉz=a+iba+ib=2ib=2iIm(z)

Hence Re(z)=z+ˉz2&Im(z)=zˉz2i

So Re(z)+Im(z)=z+ˉz2+iz+iˉz2=1i2z+1+i2ˉz  and

Re(z)Im(z)=z+ˉz2(iz+iˉz2)=1+i2z+1i2ˉz

Let A=nk=1(Re(zk)Im(zk))&B=nk=1(Re(zk)+Im(zk))

So A=nk=1((1+i2)zk+(1i2)ˉzk)=(1+i2)nk=1zk+(1i2)nk=1ˉzk and

B=nk=1((1i2)zk+(1+i2)ˉzk)=(1i2)nk=1zk+(1+i2)nk=1ˉzk

So |A|=|(1+i2)nk=1zk+(1i2)nk=1ˉzk||1+i2||nk=1zk|+|1i2||nk=1ˉzk|

But |1+i2|=|1i2|=22=12 and |zk|=|ˉzk| so A22nk=1|zk|

Similarly , |B|22nk=1|zk| Thus |AB||A||B|=2(nk=1|zk|)2

Therefore |AB|(nk=1|zk|)22                                                       Q.E.D




*___________________________
The idea of Solution credit to Ravi Prakash
        and
Diego Alvariz

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