Tuesday, August 13, 2013

Complex Computations and Inequalities (1.13-15)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercises 13-15:

MathJax TeX Test Page 13. For complex x,y show||x||y|||xy|
14. If z is complex and |z|=1, compute|1+z|2+|1z|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|>|xy| is impossible. Multiply both sides by (positive) |x|+|y| to obtain |x|2|y|2>|xy|(|x|+|y|)|xy||x+y|=|x2y2| so |x2|>|x2y2|+|y2||x2|, a contradiction. The case is parallel when |y||x|.

(14) Then z=a+bi such that a2+b2=1. We see |1z|=|z1| and now |z+1|2+|z1|2=(a+1)2+b2+(a1)2+b2=2(a2+b2)+2=4.

(15) Let va,vbCk 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 cC) or at least one of va or vb is zero. () Assume |C|2=AB and va,vb0 so A,B>0. As we have seen, |BajCbj|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,vb0 and vb=cva for some complex c. We can thus manipulateB=|bj|2=|caj|2=|c|2|aj|2=|c|2A
and¯cA=¯caj¯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 |BajCbj|2=B(|C|2AB)=0. Since vb0 and thus B>0, we have |C|2=AB. 

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