I am trying to figure out what the best method is to go about finding this locus. arg ⁡ z − a z − b = θI am aware that it must be part of an arc of a circle that passes through the points a and b. The argument means that the vector ( z − a ) leads the vector ( z − b ) by θ and so by the argument must lie on an arc of a circle due to the converse of 'angles in the same segment theorem'.My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don't need to know this, how will I know what the circle looks like.Finally, if I changed the locus to arg ⁡ z − a z − b = − θWould this just be the other part of the same circle as previously or a different circle?Perhaps you could help by showing my how it would work with the question arg ⁡ z − 2 j z + 3 = π / 3

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I am trying to figure out what the best method is to go about finding this locus.  arg  ⁡                    z        −        a                    z        −        b              =  θI am aware that it must be part of an arc of a circle that passes through the points   a and   b. The argument means that the vector   (  z  −  a  ) leads the vector   (  z  −  b  ) by   θ and so by the argument must lie on an arc of a circle due to the converse of &#039;angles in the same segment theorem&#039;.My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don&#039;t need to know this, how will I know what the circle looks like.Finally, if I changed the locus to  arg  ⁡                    z        −        a                    z        −        b              =  −  θWould this just be the other part of the same circle as previously or a different circle?Perhaps you could help by showing my how it would work with the question  arg  ⁡            z      −      2      j              z      +      3        =      π        /    3

Percyaehyq · Accepted Answer

Since you are aware that the locus would be a circle, let me just proceed to how you can get the radius of the circle. Let   A,   B and   C be the points representing the complex numbers   a,   b and   z. Then, in   Δ  A  B  C,   ∠  C  =  θ. By applying the sine rule, i.e.                    sin        ⁡        A                    B        C              =                    sin        ⁡        B                    A        C              =                    sin        ⁡        C                    A        B              =  2  Rwhere   R is the circumradius of   Δ  A  B  C, we get  R  =            sin      ⁡      C              2      A      B        =            sin      ⁡      θ              2              |            a      −      b              |              .If   arg  ⁡      (                  z        −        a                    z        −        b              )    =  −  θ, then the resulting circle will be the reflection of the circle mentioned before about the line   A  B. This happens because a change from   +  θ to   −  θ only changes the sense of rotation. I leave it up to you to verify this.EDIT: The locus will not be a circle, but only a part of the circle. Moreover, the curve will have poles at   z  =  a and   z  =  b.

I am trying to figure out what the best method is to go about finding this locus. arg &#x2

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