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2Arg(z+1) = arg(z) (1 Viewer)

uart

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Note that the lines from O to 1, from O to z and from z to z+1 form a parallelogram rhombus. The result follows immediately from the parellelogram rhombus property that the diagonals bisect the corner angles.
 
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Luukas.2

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This question was extremely easy under the old syllabus, as the following circle geometry theorem was part of the course.

If AB is a chord of a circle, centred at O, then the angle at the centre of the circle standing on that arc is twice the angle at the circumference on the same arc.

In other words, if C is any point on major arc AB of the circle, then angle AOB is exactly twice angle ACB.

Applied to this case, arg (z) corresponds to angle AOB and arg (z + 1) corresponds to angle ACB, with A being at z = 1 and B representing z.

Does this question come from an older source? The question can be done without the circle geometry, but using circle geometry is the easiest approach.
 

Drongoski

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Note that the lines from O to 1, from O to z and from z to z+1 form a parallelogram. The result follows immediately from the parellelogram property that the diagonals bisect the corner angles.
This reasoning is incorrect. You should point out that the quadrilateral so formed is a rhombus (all 4 sides equal, in this case = 1), whose diagonals bisect the angles of the rhombus. In general, diagonals of a parallelogram do not bisect their incident angles.
 
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uart

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Yes you're correct, you need to show a rhombus specifically for the angle bisect property. Sorry, I deleted that quoted post just before reading your reply and was about to acknowlege the correction.
 

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