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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > identify the region |z-2|<8&|z-1|<|z-5|
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identify the region |z-2|<8&|z-1|<|z-5|

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first we have to change inequality sign to equality sign.then |z-2|=8 represent a circle whose center lies on real axis in argand plane at (2,0) with radius 8.in complex no. |z-q| represent distance b/w two points.so when we interchange inequality with equality in second equation we get |z-1|=|z-5|.this...
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first we have to change inequality sign to equality sign.then |z-2|=8 represent a circle whose center lies on real axis in argand plane at (2,0) with radius 8.in complex no. |z-q| represent distance b/w two points.so when we interchange inequality with equality in second equation we get |z-1|=|z-5|.this says that distance of point z from pt. (-1,0) and from (-5,0) equal so the locus of z is a straight line which is perpendicular bisector of pt.(-1,0) &(-5,0) this is passes through mid point of these two points so the mid pt. is (-3,0) . this line also cuts the circle for identify the region we take any point (0,0) and put put this in both the equation if it satisfy both the inequality then the region is lies right side of the line . read less
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|z-2| < 8, let z = a + bi, with i = sqrt(-1) | (a + bi) - 2| < 8 | (a-2) + bi| < 8 sqrt < 8 ( a - 2) ^2 + b ^2 < 64 This region describes a circle of radius 8 centered at (2,0) in the complex plane. |z-1| < |z-5| (a-1)^2 + b^2 < (a-5)^2 + b^2 (a - 1) ^ 2 - (a-5)^2 < 0 4(2a - 6)...
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|z-2| < 8, let z = a + bi, with i = sqrt(-1) | (a + bi) - 2| < 8 | (a-2) + bi| < 8 sqrt [ (a-2) ^ 2 + b ^ 2 ] < 8 ( a - 2) ^2 + b ^2 < 64 This region describes a circle of radius 8 centered at (2,0) in the complex plane. |z-1| < |z-5| (a-1)^2 + b^2 < (a-5)^2 + b^2 (a - 1) ^ 2 - (a-5)^2 < 0 4(2a - 6) < 0 (a-3) < 0 a<3 The region described is the entire complex plane with real part less than 3. Hope this helps Cheers read less
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Sandeep 9 hours ago first we have to change inequality sign to equality sign.then |z-2|=8 represent a circle whose center lies on real axis in argand plane at (2,0) with radius 8.in complex no. |z-q| represent distance b/w two points.so when we interchange inequality with equality in second equation...
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Sandeep 9 hours ago first we have to change inequality sign to equality sign.then |z-2|=8 represent a circle whose center lies on real axis in argand plane at (2,0) with radius 8.in complex no. |z-q| represent distance b/w two points.so when we interchange inequality with equality in second equation we get |z-1|=|z-5|.this says that distance of point z from pt. (-1,0) and from (-5,0) equal so the locus of z is a straight line which is perpendicular bisector of pt.(-1,0) &(-5,0) this is passes through mid point of these two points so the mid pt. is (-3,0) . this line also cuts the circle for identify the region we take any point (0,0) and put put this in both the equation if it satisfy both the inequality then the region is lies right side of the line . read less
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

Circle with center (2,0) and radius 8.
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

For the other one, distance of z from 1 should be less than distance from 5.
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SME in Mathematics & Statistics

|z-2|<8 ? (x -- 2)2 + y2 < 64 The area (x -- 2)2 + y2 < 64 is interior of the circle (x -- 2)2 + y2 = 64. The circle (x -- 2)2 + y2 = 64 has centre (2, 0) and radius 8. |z-1|<|z-5| ? (x -- 1)2 < (x -- 5)2 ? 1 -- 2x < 25 -- 10x ? 8x < 24 ? x < 3 The area x < 3 is marked to the left of the straight...
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|z-2|<8 ? (x – 2)2 + y2 < 64 The area (x – 2)2 + y2 < 64 is interior of the circle (x – 2)2 + y2 = 64. The circle (x – 2)2 + y2 = 64 has centre (2, 0) and radius 8. |z-1|<|z-5| ? (x – 1)2 < (x – 5)2 ? 1 – 2x < 25 – 10x ? 8x < 24 ? x < 3 The area x < 3 is marked to the left of the straight line x = 3. The line x = 3 is parallel to y axis intersecting x axis at (3, 0). The resultant area will be the common intersection zone of the above two areas. read less
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