For complex numbers `z and w`, prove that `|z|^2w- |w|^2 z = z - w`, if and only if `z = w` or `z barw = 1`.

Explanation

Given `|z|^2w-|w|^2z=z-w`
`rArr z bar z w-wbarw z=z-w " "[because |z|^2=zbarz 1….(1)`
Taking modules of both sides ,we get
`rArr" " |zw||barz-bar w|=|z- bar w|`
`rArr" " |zw||barz-barw|=|bar(z-w)| " "[therefore |z| = | bar z|]`
`rArr" "|barz-barw|(|zw|-1)=0`
`rArr" " |bar(z-w)|=0 or |zw|-1=0 `
`rArr" " |z-w|=0 or |zw|=1`
`rArr" " z-w=0 or |zw|=1`
`rArr z=w or |zw|=1 `
Now suppose `z ne w`
Then |zw|=1 or |z||w|=1
`rArr |z|=1/|w|=r`
`Let " "z=e^(i phi) and w=1/re^(iphi)`
On putting these values in Eq.(i) we get
`r^2(1/re^(iphi))-1/r(re^(itheta))=re^(iphi)-1/re^(iphi)`
`rArr (r+1/r)^re^(iphi)=(r+1/r)^re^(iphi)`
`rArr e^(iphi)=e^(iphi ) rArr phi= theta`
Therefore , `z=re^(i phi) and w =1/re^(iphi)`
`rArr zbarw=re^(iphi).1/re^(-itheta)=1`
If and only if mean we have to prove the relation in both directions .m
Consversely
Assuming that `z =w or z bar w =1 `
If z = w then lt brgt LHS `= zbarz w-barw z=|z|^2.z - |w|^2.z`
`= |z|^2.z -|z|^2 . z =0 `
and `RHS = z-w=0 `
If `zw=1 , " then " bar(zw) = 1 bar w .1`
`LHS = bar (zz) - bar w = z=barz .1 - bar .1 `
`=barz - bar w = z - bar w =0 = RHS`
Hence proved
Alternate solution
We have `|z|^2w-|w|^2z=z-w `
`iff |z|^2 w-|w|^2 z-z + w=0 `
`iff (|z|^2 +1)w-(|w|^2+1)z`
`iff z/w=(|z|^2+1)/(|w|^2 +1)`
` therefore z/w ` is purely real.
`iff (barz)/(barw)=(z)/(w) rArr zbarw = bar w `
Again `|z|^2w-|w|^2 z=z - w `
`iff z.barz - w.barw z= z- w `
`iff z(barz w- 1) - w (zbar w - 1)=0`
`iff (z-w)(z bar w -1)=0 " " ["from Eq. (i) "]`
`iff z= w or zbarw = 1`
Therefore `|z|^2w- |w|^2 z=z - w " if and only if " z=w or z barw = 1 `