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  • A point moves in the complex plane such that <img alt="\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{4}" src="https://entrancecorner.codecogs.com/gif.latex?%5Carg%20%5Cleft%28%5Cfrac%7Bz

A point z moves in the complex plane such that \arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{4}, then the minimum value of |z-9 \sqrt{2}-2 i|^{2} is equal to_______.
 

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The locus of z is a circular arc with centre (0,2) radius 2\sqrt{2} as drawing figure
Minimum distance of z from \left ( 9\sqrt{2},2 \right ) will be along the normal (passing through centre) distance between (0,2) & \left ( 9\sqrt{2},2 \right )
= \sqrt{162+0}= 9\sqrt{2}\, units
distance between circle & \left (9\sqrt{2} ,2 \right )
= d-r= 9\sqrt{2}-2\sqrt{2}= 7\sqrt{2}\: units
d_{min}^{2}= \left ( 7\sqrt{2} \right )^{2}= 98

Posted by

Kuldeep Maurya

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