Prove if <mrow class="MJX-TeXAtom-ORD"> | </mrow> z <mrow class="MJX-TeXAtom-ORD">

encamineu2cki

encamineu2cki

Answered question

2022-05-08

Prove if | z | < 1 and | w | < 1, then | 1 z w | 0 and | z w 1 z w | < 1
Given that | 1 z w | 2 | z w | 2 = ( 1 | z | 2 ) ( 1 | w | 2 )
I think the first part can be proven by saying | 1 z w | = 0 if and only if z w = 1.

And given the conditions that cannot be true. However I don't know if this part is right.

Answer & Explanation

Abigailf91er

Abigailf91er

Beginner2022-05-09Added 13 answers

You're right! After all, since | w | = | w | < 1 and | z | < 1, then | z w | = | z | | w | < 1 = | 1 | , so that's enough.
For the second, you must equivalently show that | z w | < | 1 z w | . It suffices to show that
| z w | 2 < | 1 z w | 2 ,
or equivalently that
| 1 z w | 2 | z w | 2 > 0.
Now use the given equation, together with the fact that | z | 2 < 1 and | w | 2 < 1
vilitatelp014

vilitatelp014

Beginner2022-05-10Added 6 answers

| z w | | 1 w ¯ z | < 1
iff
| z w | < | 1 w ¯ z |
iff
| z w | 2 < | 1 w ¯ z | 2 .
Therefore, we must check if
( z w ) ( z ¯ w ¯ ) < ( 1 w ¯ z ) ( 1 w z ¯ ) .
In other words, if
| z | 2 z w ¯ z ¯ w + | w | 2 < 1 w ¯ z z ¯ w + | z | 2 | w | 2 .
Therefore, the statement becomes, if 0 a , b < 1, then
a + b < 1 + a b .
This, however, is obvious since
0 < 1 a b + a b = ( 1 a ) ( 1 b )
is true. Now, reverse all the steps to get a proof.

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