a r g ( z − a z − b ) = cMy understanding is as follows. The angle c between the lines za and zb is constant. za and zb meet at z and the angle between the lines perpendicular to za and zb is 2c, which is only the case if z lies on a circle passing through z, a and b with centre where the lines perpendicular to za and zb intersect. So the locus is an arc of a circle through a and b except a and b (because a r g ( z − a z − b ) is undefined there). Is this reasoning correct?Is there another way the statement could this be expressed (e.g. in terms of moduli)? Also, what is an efficient way of finding the centre of the circle?

Question

a  r  g  (            z      −      a              z      −      b        )  =  cMy understanding is as follows. The angle c between the lines za and zb is constant. za and zb meet at z and the angle between the lines perpendicular to za and zb is 2c, which is only the case if z lies on a circle passing through z, a and b with centre where the lines perpendicular to za and zb intersect. So the locus is an arc of a circle through a and b except a and b (because   a  r  g  (            z      −      a              z      −      b        ) is undefined there). Is this reasoning correct?Is there another way the statement could this be expressed (e.g. in terms of moduli)? Also, what is an efficient way of finding the centre of the circle?

pyphekam · Accepted Answer

Write   z  =  x  +  i  y and multiply top and bottom of the given fraction by the conjugate of   z  −  b. Identify the real and imaginary parts as   p and   q, say, so that you have  arg  ⁡  (  p  +  i  q  )  =  cNow, depending on the value of the angle   c, you can form an equation for   x and   y which will reduce to the equation of a circle.We can identify that it is a part circle by considering which quadrant   c lies in. For example if c is acute, we would have the additional requirement that   a  &amp;gt;  0 and   b  &amp;gt;  0, and similarly for other angles.

a r g ( z − a </mrow>

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