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 D7,3
Let 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.