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If denotes the number of solutions of |1-i|x=2x and = ( |z| (z) ), where z= 4(1+i)4 [ 1- i +i+ -i1+ i ], i= -1, then the distance of the point ( , ) from the line 4x-3y=7 is .
3
( 2 )x=2x ⇒ ( 1 2 )x=1 ⇒ x=0 only ⇒ =1
1- i=-i ( +i ) and -i=-i (1+ i ) ⇒ bracket =-i-i=-2i
(1+i)4=(2i)2=-4 ⇒ z= 4(-4)(-2i)=2 i
|z|=2 , (z)= 2 ⇒ =4
Distance of (1, 4) from 4x-3y-7=0: |4-12-7|5=3
Answer: 3
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