If argument of z − z 1 z − z 2 is π 4 , find the locus of z. z 1 = 2 + 3 i z 2 = 6 + 9 iApproach: I tried to solve the equation using diagram, basically plotting the points on the Argand plane. What I got is a circle with center 7 + 4 i and a radius of 26 units. The two complex numbers given lie on this circle, and form a chord. Any point lying on the major arc of this chord satisfies the condition.How exactly would I represent this as a locus of the point? And is there any other method that I can use that does not involve a diagram?

Question

If argument of             z      −              z        1                    z      −              z        2             is       π    4  , find the locus of   z.      z    1    =  2  +  3  i      z    2    =  6  +  9  iApproach: I tried to solve the equation using diagram, basically plotting the points on the Argand plane. What I got is a circle with center   7  +  4  i and a radius of       26   units. The two complex numbers given lie on this circle, and form a chord. Any point lying on the major arc of this chord satisfies the condition.How exactly would I represent this as a locus of the point? And is there any other method that I can use that does not involve a diagram?

Jenna Farmer · Accepted Answer

the angle subtended by the chord       z    1        z    2   at the center is   2  π      /    4  =  π      /    2 so the radius is                     |                    z        1            −              z        2                    |                    2        =      26   the center of the chord is   4  +  3  i you add or subtract                     −        6        +        4        i            2       so that you will get two centers. the two centres,       z    1   and       z    2   form a square of side       26    .

Savanah Boone · Accepted Answer

Put     z  =  x  +  i  y    ,      x  ,  y  ∈      R    , so            z      −      2      −      3      i              z      −      6      −      9      i        =            (      x      −      2      )      +      (      y      −      3      )      i              (      x      −      6      )      +      (      y      −      9      )      i        ⋅            (      x      −      6      )      −      (      y      −      9      )      i              (      x      −      6      )      −      (      y      −      9      )      i        =  =            (      x      −      2      )      (      x      −      6      )      +      (      y      −      3      )      (      y      −      9      )              (      x      −      6              )        2            +      (      y      −      9              )        2              +            (      x      −      6      )      (      y      −      3      )      −      (      x      −      2      )      (      y      −      9      )              (      x      −      6              )        2            +      (      y      −      9              )        2              iBy the given data, it must be that the real and imaginary parts are identical, and thus  (  x  −  2  )  (  x  −  6  )  +  (  y  −  3  )  (  y  −  9  )  =  (  x  −  6  )  (  y  −  3  )  −  (  x  −  2  )  (  y  −  9  )    ⟺      ⟺        x    2    −  14  x  +      y    2    −  8  y  −  26  =  0Complete squares, make some algebraic hokus pokus and get a circle.