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Locus of arg(z/(z-1+2i)) = p/3
By Sarah Shales (t382) on June 8, 1998:

Please could you help me with explaining what the locus of
arg (z/(z-1+2i)) = p/3 looks like ?

I know it is an arc of a circle, but which arc ?

Help

By Stuart White (saw30) on June 10, 1998:

Dear Sarah,

As arg (a×b) = arg a arg b the equation becomes:
arg z - arg (z - ( 1-2i ) ) = p/3

The rest of the answer gets somewhat geometrical, so look at the diagram.

Draw a triangle OAB, with O at the origin and B at (1,-2). Let A represent z. Draw lines OX, Y'BY parallel to the x-axis as shown.

Then arg z = XOA (where XOA represents the angle between XO and OA) & arg (z - (1 - 2i ) ) = YBA. Therefore the equations can be written
XOA - YBA = p/3

As BOX = Y'BO we can write
(BOA - Y'BO) - YBA = p/3
BOA - (Y'BO + YBA) = p/3

By angles on a straight line (Y'OY) this becomes:
BOA - (p - ABO) = p/3

and by angles in a triangle:
OAB = p/3

(Note I have been careful to ensure all angles are described in an anticlockwise sense)

So the angle OAB is constant, and as all angles subtended by the same arc of a circle are equal A lies on an arc of a circle connecting O to B, such that the size of the angle |OAB| is p/3. There are two such arcs, one below O & B and one above.

One arc will correspond to the p/3 and the other -p/3, so how do we find out which is which? Well here's a rather crude way:

For A to be above OB clearly YBA >= XOA (measured as arguments i.e. in an anticlockwise sense). Therefore arg z - arg (z- (1 - 2i) ) < 0 so does not equal p/3.

So A must lie below OB. So we have determined which arc the locus of z lies on.

I hope this helps - please get back to me if it makes no sense.

Yours,

Stuart