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Question
If |z| = 3, show that 7 ≤ |z + 6 – 8i| ≤ 13
Solution
Given |z| = 3
|z + 6 – 8i| ≤ |z| + |6 – 8i|
= `3 + sqrt(6^2 + 8^2)`
= `3 + sqrt(100)`
= 3 + 10 = 13
∴ |z + 6 – 8i| ≤ 13 ........(1)
|z + 6 – 8i| ≥ ||z| – |– 6 + 8i||
= |3 – 10|
= |– 7|
= 7
∴ |z + 6 – 8i| ≥ 7 .........(2)
From 1 and 2
we get 7 ≤ |z + 6 – 8i| ≤ 13
Hence proved.
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