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If |z|<=1,|w|<=1, then show that |z- w|...

If ` |z|<=1,|w|<=1`, then show that `|z- w|^2<=(|z|-|w|)^2+(argz-argw)^2`

Text Solution

Verified by Experts

Let `z =|z|e^(ialpha) and w =|w|e^(ibeta)`
Now, `|z -w|^(2) =|z|^(2) +|w|^(2) -2Re (zbarw)`
`= (|z| -|w|)^(2) -2|z||w| -2Re(|z||w|e^(i(alpha -beta)))`
`=(|z| -|w|)^(2) +|z||w|(2-2cos (alpha- beta))`
` lt (|z| - |w|)^(2) + 4sin^(2) ((alpha-beta)/(2)) " "(because |z| le 1,|w| le1)`
`le (|z| -|w|)^(2) + 4((alpha - beta)/(2))^(2) " " (because sin theta lt theta "for" theta in (0,pi//2))`
`= (|z| -|w|)^(2) + 4(alpha - beta)^(2)`
`= (|z| -|w|)^(2) + (arg a -arg w)^(2)`
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