Question 1156628
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Find all complex numbers z such that z^2=z with line on top or complex conjugate
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<pre>
It is better and much easier to analyse and to solve this problem in polar trigonometric form.


If z^2 = z complex cojugate,  then, firstly, the modulus of z is equal to 1.

In other words, z lies on the unit circle in a coordinate plane.


Next, if the argument of z is polar angle {{{alpha}}},  then the polar angle of z^2 is  {{{2*alpha}}},

while the polar angle of (z conjugate) is  {{{-alpha}}}  or  {{{2pi - alpha}}}.


So, we get the equation for the polar angle


    <U>Case 1</U>.  {{{2*alpha}}} = -{{{alpha}}},  which implies  {{{3alpha}}} = 0,  or  {{{alpha}}} = 0.   Then the solution is z = 1.


OR


    <U>Case 2</U>.  {{{2*alpha}}} = {{{2pi- alpha}}},  which implies  {{{3alpha}}} = {{{2pi}}}.    Hence,  {{{alpha}}} = {{{2pi/3}}}.


<U>ANSWER</U>.  There are TWO solutions.  One solution is z = 1.

          The other solution is  z = {{{cos(2pi/3) + i*sin(2pi/3)}}} = {{{(-1/2) + i*(sqrt(3)/2)}}}.
</pre>

Solved.