Solve #|z-w|=|z+w| # ?

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Answer 1 · 2018-02-03T18:56:11

See below.

From #|z-w|=|z+w| # we conclude

#(z-w)bar((z-w)) = (z-w)(barz-barw) = z barz + w bar w-z bar w-w barz = z bar z+w bar w+z bar w + w bar z# or

#z barz + w bar w-z bar w-w barz = z bar z+w bar w+z bar w + w bar z# or

#2(z bar w +w bar z) = 0#

now making

#z = alpha e^(i phi) w# and substituting

#alpha e^(-i phi)w bar w+alpha e^(i phi) w bar w = 0# or

#alpha(e^(i phi)+e^(-iphi))w bar w = 0# but #w bar w > 0# and #alpha gt 0#

then

#e^(i phi)+e^(-iphi) = 0 rArr phi=pi/2#

This fact implies on

#z =alpha e^(i pi/2) w =alpha i w#