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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > Find the complex number z |z+2| > |z-3|
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Find the complex number z |z+2| > |z-3|

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Z could be any number
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Simply Re(z) is less than 1/2. Either use z = x + iy and solve. Or simply use the concept of perpendicular bisector.
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Re(z) is less than 1/2. Either substitute z = x + iy and solve. Or use the concept of perpendicular bisector.
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Let z=x+iy |z+2| = sqrt ((x+2)^2+y^2) squaring both sides |z+2|^2 = (x+2)^2+y^2 Similarly |z-3|^2 = (x-3)^2+y^2 To find, when (x+2)^2 > (x-3)^2 as y^2 gets cancelled. Simplifying, 4x+4>-6x+9 10x>5 x>1/2 and y can take any value.
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Lecturer in Mathematics in IIT level, 22 years experience and qualification M.Sc(Maths)

x> 1/2.
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Since z=x + yi is the kind of complex number we are looking out for ....therefore we substitute this in the given inequality. | (x+yi) + 2 | > | (x+yi) - 3 | | (x+2) + yi | > | (x-3) + yi | But | x + yi | = sqrt (x^2 - y^2), applying this we get sqrt > sqrt squaring both sides we get (x+2)...
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Since z=x + yi is the kind of complex number we are looking out for ....therefore we substitute this in the given inequality. | (x+yi) + 2 | > | (x+yi) - 3 | | (x+2) + yi | > | (x-3) + yi | But | x + yi | = sqrt (x^2 - y^2), applying this we get sqrt [ (x+2) ^2 - y^2 ] > sqrt [ (x-3)^2 - y^2 ] squaring both sides we get (x+2) ^2 - y^2 > (x-3)^2 - y^2 (x+2) ^ 2 - (x-3) ^ 2 > 0 (x + 2 + x - 3) (x + 2 - x + 3) > 0, since a^2 - b^2 = (a+b)(a-b) (2x - 1)(5) > 0 (2x - 1) > 0 x > (1/2) Hence all complex numbers of the form z=x + yi, with x > 0.5 satisfy the given inequality, note that y can also take negative real values and I have therefore ignored the case z=x - yi. Geometrically this means , z is such a complex number such that its distance from the point (-2,0) is greater than its distance from the point (3,0). If you should draw a number line and locate these points, its clear that all z with x > 0.5 satisfy the condition. Such numbers lie to the right of the vertical line passing through x = 0.5. read less
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|z+2| it is the distance of pt. z from pt (-2,0) in the complex plane and |z-3| represent the distance of pt. z from (3,0) in the complex plane. now you know what is mode of a complex no. represent . to solve this type of question you have to first identify the region to find the region you have to...
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|z+2| it is the distance of pt. z from pt (-2,0) in the complex plane and |z-3| represent the distance of pt. z from (3,0) in the complex plane. now you know what is mode of a complex no. represent . to solve this type of question you have to first identify the region to find the region you have to replace inequality sign with equality sign first . i.e |z+2|=|z-3| say it equation -(a ) if you sea in your notes this is the equation bisector of fixed pt.(-2,0)and (3,0) so the locus of z is a line which is perpendicular bisector of line joining pt (-2,0) &(3,0) so we get a line from equation (a) now we have to find z(in other words we have to find region where z lies because inequalities always satisfies a region) region in this question is the values of z which satisfies the inequality.first perpendicular bisector is passes through (-0.5,y) in the complex plane.so the z is any complex no .whose real part is fixed i.e -0.5 but imaginary part can take any value on imaginary axis say k where k belongs to (-infinity ,+infinity). i hope you understand .please sea your notes also while study this. read less
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D=WT/VOL
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A true teacher of Science and Math

I dint get answer from anyone, two complex no never compare
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I x +iy +2 I > I x +iy -3 I Sqrt((x+2)^2 +y^2) > Sqrt((x-3)^2 +y^2) I After simplification you can obtain x > 1/2
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