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If |z| = 2 show that the 8 ≤ |z + 6 + 8i| ≤ 12 - Mathematics

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प्रश्न

If |z| = 2 show that the 8 ≤ |z + 6 + 8i| ≤ 12

बेरीज

उत्तर

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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Modulus of a Complex Number
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 2: Complex Numbers - Exercise 2.5 [पृष्ठ ७२]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 2 Complex Numbers
Exercise 2.5 | Q 6 | पृष्ठ ७२
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