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If z is a complex number lying in the fo...

If `z` is a complex number lying in the fourth quadrant of Argand plane and `|[(kz)/(k+1)]+2i|>sqrt(2)` for all real value of`k(k!=-1),` then range of `"arg"(z)` is `(pi/8,0)` b. `(pi/6,0)` c.`(pi/4,0)` d. none of these

Text Solution

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The given inequality is ` |(kz)/(k+1) + 2i|gt sqrt2`
`Rightarrow |z-((-2i)(k+1))/k|gt(sqrt2(k+1))/k`
Which represents the region lying outside the circle centred
`at(0,(-2(k+1))/k) and " radius " (sqrt2(k+1))/k`
let `COP =theta` so that
`sin theta= (CP)/(OC) = ((sqrt2(k+1))/k)/((2(k+1))/k)=1/sqrt2 = sinpi//4`
` theta= pi/4`
Since z lies in the `IV^("th")` quadrant, hence
` arg z "in" ( - pi/4, 0)`
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