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Complex numbers z satisfy the equaiton |z-(4//z)|=2 Locus of z if |z-z_(1)| = |z-z_(2)|, where z_(1)and z_(2) are complex numbers withthe greatest and the least moduli, is |
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Answer» lineparallel to the real axis 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` `rArrsqrt(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 `rArrz=pm(sqrt(5)+1)` similarly for |z| =sqrt(5)-1` , we have `z= pm(sqrt(5)-1)` |
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