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

Explanation

Let `z=r_1 (cos theta_1 + I sin theta_1)` + `w= r_2( cos theta_2)+ is (r_1 sin theta_2)+ I (r_1 sin theta_1-r_2 sin theta _2)`
we have `|z|= r_1 ,|w| =r_2 arg (z)=theta_1 ` and arg `(w) = theta_2`
Given `|z| le 1, |w| = r_2 and (z)= theta_2 le 1 `
Now
`z-w=(r_1 cos theta_1-r_2 cos theta_2)+ i ( r_1 sin theta _1 - r_2 sin theta_2)`
`rArr |z-w|^2=(r_1 cos theta_1 -r_2 cos theta_2 )^2+(r_1 sin theta_1- r_2 sin theta_1-r_2 sin theta_2 )^2`
`=r_1^2 cos ^2 theta_1 + r_2 ^2cos^2 theta_2- 2r_1 r_2 cos theta_1c ostheta_2+r_1^2cos ^2 theta_1+ sin^2theta_1+r_2^2sin^2theta_2-2r_1r_2sin theta_1 sin theta_2`
`=r_1^2(cos^2 theta_1+ sin ^2 theta_1)+r_2^2(cos^2 theta_1 + sin ^2theta_2)-2r_1r_2(cos theta_1 cos theta_2 + sin theta_1 sin theta_2)`
`=r_1^2 +r_2^2-2r_1r_2cos (theta_1- theta_2)`
`=(r_1 -r_2)^2 + 4r_1r_2[ 1- cos ( theta_1 - theta _2)]`
` = (r_1 - r_2)^2 + 4r_1r_2 sin ^2 (( theta_1 - theta_2)/(2))le |r_1-r_2|^2+ 4 | sin ((theta _1 - theta_2)|^2 " "[ because r_1.r_2 le 1 ]`
and `|sin theta| le | theta|, AA theta in R `
Therefore `|z-w|^2 le | r_1 - r_2|^2 + 4 |( theta_1 - theta _2)/2|^2`
` le |r_1 -r_2|^2+ | theta _1 theta_2|^2`
`le |r_1-r_2|^2 + | theta _1 - theta_2|^2`
`rArr | z-w|^2 le ( |z| -|w|^2+ |arg z- arg w)^2`
Alternate Solution
`|z-w|^2= |z|^2 + |w|^2 -2 |z||w| cos (arg z- arg w)`
`= |z|^2+ |w|-2 |z||w| cos (arg z- arg w) `
`= (|z| - |w|)^2+ 2|w|.2 sin ^2((arg z- arg w )/2) `
`therefore | z-w|^2 le (|z| - |w|)^2 + 4.1.1((arg z- arg w )/2)^2 " "[ because sin theta le theta ]`
`rArr |z-w|^2 le (|z| -|w|)^2 + (arg z - arg w)^2`