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Let complex numbers `alpha and 1/alpha` lies on circle `(x-x_0)^2(y-y_0)^2=r^2 and (x-x_0)^2+(y-y_0)^2=4r^2` respectively. If `z_0=x_0+iy_0` satisfies the equation `2|z_0|^2=r^2+2` then `|alpha|` is equal to (a) `1/sqrt2` (b) `1/2` (c) `1/sqrt7` (d) `1/3`A. `1/(sqrt(2))`B. `1/2`C. `1/sqrt(7)`D. `1/3` |
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Answer» Correct Answer - C Formula used `|z|^2=z. bar z` and `|z_1=z_2|^2=(z_1-z_2)(barz_1-z_2)` `=|z_1|^2-z_1barz_2-z_2bar z_1+|z|^2` Here `(x-x_0)^2+(y-y_0)^2` and `(x-x_0)^2+(y-y_0)^2=4r^2` since `alpha and 1/alpha` lies on first and second respectively `therefore " "|alpha-z_0|^2=r^2 and |(1)/(bar alpha -z_0)|^2=4r^2` `rArr (alpha-z_0)(baralpha-alpha _0)=r^2` `rArr |alpha^2|-z_0 bar alpha- bar z_0 alpha+|z_0|^2=r^3` and `|1/(alpha)-z_0|^2=4r^2` `rArr 1/alpha-z_0=4r^2` `rArr 1/(|alpha|^2)-(z)/alpha-(barz_0)/(baralpha)+|z_0|^2=4r^2` Since `|alpha|=alpha.alpha` `1/(|alpha|^2)-(z_0.baralpha)/(|alpha|^2)-barz_0/(|alpha|^2).alpha+|z_0|^2=4r^2` `rArr 1-z_0baralpha-barz_0alpha+|alpha|^2|z_0|^2=4r^2|alpha|^2` On subtracting Eqs. (i) and (ii) ,we get `(|alpha|^2-1)(1-|z_0|^2)=r^2(1-4|alpha|^2)` `rArr(|alpha|^2-1)(1-|z_0|^2)=r^2(1-4|alpha|^2)` `rArr (|alpha|^2-1)(1-(r^2+2)/(2))=r^2(1-4|alpha|^2) ` Given `|z_0|^2=(r^2+2)/(2)` `rArr (|alpha|^2-1).((-r^2)/2)=r^2(1-4|alpha|^2)` `rArr |alpha|^2-1=-2+8|alpha|^2` `rArr |alpha|^2-1=-2+8|alpha|^2` `rArr " "7|alpha|^2=1` `therefore |alpha|=1 //sqrt(2)` |
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