grimie said:
can you have a look at this for me?
Sketch the region on the argand diagram that satisfies the inequality:
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zz + 2(z+z) (lesser than or equal to) 0
do u understand that the second and fourth "z s" are conjuagtes? sorry - it is really hard to type maths quesitons on a normal comp
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z=x-iy
z=x+iy
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zz= x^2 + y^2
2(z+congZ)= 2(x+iy) + 2(x-iy) = 4x
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zz + 2(congZ) = x^2 + y^2 + 4x <= 0
you should be able to graph and shade that.
Wait I'll try
(x+2)^2 + y^2 - 4 <=0
(x+2)^2 + y^2 <=4
therefore circle radius 2, center (-2, 0)
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