Complex numbers z satisfy the equaiton `|z-(4//z)|=2`
The difference between the least and the greatest moduli of complex number is

Options

A. 2
B. 4
C. 1
D. 3

Answer

A

Explanation

We have `||z|-|4/z|| le |z-4/5|=2`
`rArr -2 le |z| - 4/|z| le 2 `
`rArr |z|^(2) +2 |z| -4 ge0 ` and `|z|^(2) -2|z|-4 le 0`
` rArr (|z|+1)^(2)-5 le 0 ` and `(|z|-1)^(2) le 5`
`rArr (|z|+1 + sqrt(5)) (|z|+1-sqrt(5)) ge0`
`rArr |z| ge sqrt(5) -1`
and `(|z|- 1+sqrt(5)) xx(|z|-1-sqrt(5)) le 0`
`rArr sqrt(5)-1 le |z| ge sqrt(5)-1`
and `(|z| -1 + sqrt(5)) (|z| -1-sqrt(5)) le0`
`rArr sqrt(5)-1 le |z| le sqrt(5)+1`
`rArr sqrt(5) -1 le |z| le sqrt(5)+1 `
Hence, the least modulus is `sqrt(5)-1` and the greatest modulus is `sqrt(5) +1`. also,
`|z|=sqrt(5) +1 `
`rArr 4/(|z|) =sqrt(5)-1`
Now `4/z =(4bar(z))/(|z|^(2))`
Hence, `4//z` lies in the direction of `bar(z)`
`|z-4/z|=PR=2 ` (given )
We have
`OP =sqrt(5) +1 ` and `OR =sqrt(5)-1`
`rArr cos 2 theta=(OP^(2)+OR^(2)+PR^(2))/(2OP.OR)`
`=((sqrt(5)+1)^(2)+(sqrt(5)-1)^(2)-4)/(2(5-1))=1`
`rArr 2 theta=0, 2pi`
`rArr theta=0, pi`
`rArr z ` is purely real
`rArr z=pm(sqrt(5)+1)`
similarly for |z| =sqrt(5)-1` , we have `z= pm(sqrt(5)-1)`