Please could you help me with explaining what the locus of

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 O A B, with O at the origin and B at (1,-2). Let A represent z. Draw lines O X, Y¢B Y parallel to the x-axis as shown.

Then arg z = X O A (where X O A represents the angle between X O and O A) and arg (z - (1 - 2i ) ) = Y B A. Therefore the equations can be written

X O A - Y B A = p/3

As B O X = Y¢B O we can write

(B O A - Y¢B O) - Y B A = p/3

B O A - (Y¢B O + Y B A) = p/3

By angles on a straight line (Y¢O Y) this becomes:

B O A - (p- A B O) = p/3

and by angles in a triangle:

O A B = p/3

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

So the angle O A B 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 |O A B| is p/3. There are two such arcs, one below O and 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 O B clearly Y B A ³ X O A (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 O B. 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