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a
1+2
b
2+2
c
3+1
d
5+1
answer is D.
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Detailed Solution
We have |z|=|z−4z+4z|≤|z−4z|+4|z| ⇒ |z| ≤2+4|z| ⇒ |z|2 ≤ 2|z|+4 ⇒ (|z|−1)2≤5⇒ |z|−1 ≤ 5 ⇒ |z| ≤ 5+1Also, for z=5+1, |z−4z|=2Therefore, the greatest value of |z| is 5+1