Let z be a complex number satisfying |z – 5i|≤1 such that arg(z) is minimum. Then z is equal to
A
2√65+24i5
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B
245+2√6i5
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C
6√25+24i5
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D
i
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Solution
The correct option is A2√65+24i5 Sin∠AOC=15sin(90−θ)=15cosθ=15 OA=√OC2−CA2=√24sinθ=√245. Now the co - ordinates of A are (√24cosθ,√24sinθ)(√245,245)=√245+i(245)