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Question

# A point ′z′ moves on the curve |z−4−3i|=2 in an argand plane. The maximum and minimum values of |z| are

A
2,1
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B
6,5
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C
4,3
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D
7,3
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Solution

## The correct option is D 7,3Let w=4+3i. We can write, |z|=|(z−w)+w|. Hence by triangle inequality (|z1+z2|≤|z1|+|z2|), we can write |z|≤|z−w|+|w|. It is given in the question that, |z−w|=2 and |w|=√42+32=5.Putting the values, we get |z|≤7. Using another result of triangle inequality (∣∣|z1|−|z2|∣∣≤|z1+z2|), we can write ∣∣|z−w|−|w|∣∣≤|z−w+w|.Hence, we get |z|≥3. The minimum value is 3 and maximum value is 7.Hence, (D) is the correct option

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