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Pure Mathematics 3 question for may/june 2013 exam

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Can any one help me with this sum's part iv below please!
(I could do till iii but got badly stuck with part iv)

Q.The complex number −2 + i is denoted by u.
(i) Given that u is a root of the equation x
3 − 11x − k = 0, where k is real, find the value of k. [3]
(ii) Write down the other complex root of this equation. [1]
(iii) Find the modulus and argument of u. [2]
(iv) Sketch an Argand diagram showing the point representing u. Shade the region whose points
represent the complex numbers z satisfying both the inequalities
|z| < |z − 2| and 0 < arg(z − u) < 1/4π.
Thank u! May u be blessed for helping!:)
 
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hi.i'll explain for part (iv) only. |z| < |z-2| can be written as |z - (0,0)| < |z- (2,0)|. so make the perpendicular bisector of line joining (0,0) and (2,0) and shade the appropriate region. for arg(z-u) is same as arg(z-(-2+i)) = arg(2-i).hence draw the half required angle. hope i've clarified u a bit.
 
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Thank u it helped! :)
But one question dont we need to draw any circle as the locus?If u can please draw n show me how the diagram will be (U can draw in comp or draw in paper take pic n upload) because i m not still sure about the diagram i came with n mark scheme is really gibberish! .... Thank u very much for helping! :)
 
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