14. If z is complex and |z|=1, compute|1+z|2+|1−z|2
15. Under what conditions does equality hold in the Schwarz inequality? I.e.,|∑aj¯bj|2=∑|aj|2∑|bj|2
for complex aj,bj.
Proof: (13) Assume |x|≥|y|. Then we must show |x|−|y|>|x−y| is impossible. Multiply both sides by (positive) |x|+|y| to obtain |x|2−|y|2>|x−y|(|x|+|y|)≥|x−y||x+y|=|x2−y2| so |x2|>|x2−y2|+|y2|≥|x2|, a contradiction. The case is parallel when |y|≥|x|.
(14) Then z=a+bi such that a2+b2=1. We see |1−z|=|z−1| and now |z+1|2+|z−1|2=(a+1)2+b2+(a−1)2+b2=2(a2+b2)+2=4.
(15) Let va,vb∈Ck be the the vectors of the aj and bj. Let A=∑|aj|2,B=∑|bj|2, and C=∑aj¯bj. We claim |C|2=AB if and only if vb is associate to va (i.e. vb=cva for some c∈C) or at least one of va or vb is zero. (⇒) Assume |C|2=AB and va,vb≠0 so A,B>0. As we have seen, ∑|Baj−Cbj|2=B(AB−|C|2). When |C|2=AB we then thus have aj=CBbj for all j, and thus vb=CBva. (⇐) When either one of va or vb is zero the equality clearly holds, so assume va,vb≠0 and vb=cva for some complex c. We can thus manipulateB=∑|bj|2=∑|caj|2=|c|2∑|aj|2=|c|2A
and¯cA=¯c∑aj¯aj=∑aj¯bj=C
so for all j we haveBaj=|c|2Aaj=|c|2aj¯ajA¯aj=|caj|2¯ajA=bj¯bj¯ajA=bj¯cA=Cbj
so that ∑|Baj−Cbj|2=B(|C|2−AB)=0. Since vb≠0 and thus B>0, we have |C|2=AB. ◻
No comments:
Post a Comment