First slide
Geometry of complex numbers

If the complex number z is such that |z25|25 and |z50|=25 then the minimum value of |z|+25 is 10k  then k is equal to


|z25|25 represent interior and boundary of circle with 

centre (25, 0) and radius 25. Points z satisfying  |z25|25

and  |z50|=25  lie on arc DAC hence min {|z|}=OA=25

|z|+25=50 So, k=5

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