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Question
If |z| = 2 show that the 8 ≤ |z + 6 + 8i| ≤ 12
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Solution
Given |z| = 2
|z + 6 + 8i| = |z| + |6 + 8i|
= `2 + sqrt(6^2 + 8^2)`
= `2 + sqrt(100)`
= 2 + 10
= 12
∴ |z + 6 + 8i| ≤ 12 ........(1)
|z + 6 + 8i| ≥ ||z| – |-6 – 8i||
= |2 – 10|
= |– 8|
= 8
|z + 6 + 8i| ≥ 8 ........(2)
From 1 and 2 we get
8 ≤ |z + 6 + 8i| ≤ 12
Hence proved
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