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Very simple complex numbers question (1 Viewer)

gurmies

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(i) Using the exterior angle of the triangle inscribed, @ + arg(z+1) = arg(z-3) ===> @ = arg(z-3) - arg(z+1) = pi/3(ii) This question is quite complex and requires you to recall some circle properties. Alternatively you could use inverse trig by letting z = x+iy, which will eventually give you a circle equation after large amounts of algebra.
 

chingyloke

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for part ii. Let C(p,q) be the centre of the circle. Since centre lies on the perpendicular bisector of a chord, the the centre lies of the midpoint between x=-1 and x=3 Therefore p=1 Angle at centre is 2pi/3 (angle at centre is twice angle at circumference) Construct triangle ACB but only use half of it. tanpi/3 = 2/q 2/q=root3 qroot3 = 2 q=2/root3 rationalise denominator q=2root3/3 C(1,2root3/3)
 

untouchablecuz

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its not that we know or not; we use the property that the angle at the centre is twice the angle at the circumference to construct a triangle that will allow us to find the centre
 

chingyloke

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Thanks Guys, But sorry for sounding stupid, but how do you know the triangle touches the centre.
By the circle property. Angle at centre is twice angle at circumference subtended from same arc...because the angle 2pi/3 it must be the centre.
 

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