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Let z be a complex number satisfying |z+16| = 4|z+1| then

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|z| = 4

< |z| < 6

|z| = 5

|z| = 6

answer is A.

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Detailed Solution

$| z + 16 |^{2} = 16 | z + 1 |^{2}$

$\Rightarrow (z + 16) (z + 16) = 16 (z + 1) (z + 1)$

$\Rightarrow z z + 16 z + 16 z + 256 = 16 z z + 16 z + 16 z + 16$

$\Rightarrow z \bar{z} = 16 \Rightarrow | z |^{2} = 16 \Rightarrow | z | = 4$

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