Question 12254
 1.find the cube root of 3+isquare root 3
 Do you mean cubic root of(3 + i sqrt(3)) I guess do.
 To solve: {{{ x^3 = 3 + i sqrt(3)}}}
    Since {{{ 3 + i sqrt(3) = r  (cos w  + i sin w) }}}
    where r = sqrt(3^2+ 3) = {{{2 sqrt(3) }}}
     {{{ tan w = sqrt(3)/3 = 1/sqrt(3) }}}, i.e. w = pi/6.
  By De Mieve theorem, {{{ x = r^(1/3) (cos (pi/18 + 2* pi * k/3) 
   + i sin (pi/18 + 2* pi * k/3)) }}} ,k = 0,1,2
  Hence, {{{ x = 2^(1/3)*(3)^(1/6) (cos (pi/18) + i sin (pi/18)) }}}
   {{{ primitive cubic root, pi/18 = 20 deg }}}
  or {{{ x = 2^(1/3)(3)^(1/6) (cos ( 13 * pi/18) + i sin (13* pi/18)) }}}
  {{{ 13* pi/18 = 130 deg }}}
  or {{{ x = 2^(1/3)(3)^(1/6) (cos ( 25 *pi/18) + i sin (25 * pi/18)) }}}
    {{{ 25* pi/18 = 250 deg }}}

2.if z=x+iy ,prove that |x|+|y| less than or equal to sqareroot(2)|z|
   proof: Since {{{ (sqrt(2)*abs(z))^2 = 2*abs(z)^2 = 2*(x^2 + y^2) }}}
      and  {{{ (abs(x) + abs(y))^2 = x^2 + y^2  + 2 *abs(x) abs(y) }}}
      Consider {{{ 2*(x^2 + y^2) - (x^2 + y^2  + 2*abs(x) abs(y))  }}}
      = {{{  x^2 + y^2  - 2*abs(x)*abs(y) }}}
      = {{{ (abs(x) -abs(y))^2 >= 0 }}}
      Hence, {{{ ( abs(x) + abs(y))^2 <= 2* (abs(z))^2  }}}
      By taking sqrt, we have {{{ abs(x) + abs(y) <= sqrt(2)*abs(z) }}}
 
 **You should define what w (omega) is, I guess, it is  
 {{{ e^(2*pi*i/3) = cos (2*pi/3) + i sin (2*pi/3) }}} 
3.prove [ a+b omega+c(omega)^2] divided by [a omega+b (omega)^2+c]=(omega)^2
 Proof: {{{(a + b *w  + c *w^2)/ (a *w + b *w^2  + c) }}}
        = {{{ w^2 *(a + b *w  + c *w^2)/ (w^2*(a *w + b *w^2  + c)) }}}
        = {{{ w^2 *(a + b *w  + c *w^2)/ (a *w^3 + b *w^4  + c*w^2)) }}}
       [ Since {{{ w^3 = 1 }}}, {{{w^4 = w }}} ]
        = {{{ w^2 *(a + b *w  + c *w^2)/ (a + b *w  + c*w^2)) }}}   
        = {{{ w^2 }}} [ {{{ (a + b *w  + c *w^2)}}} is cancelled ]

4.prove 1+(omega)^n+(omega)^2n=0 when n=2,4
  Proof: Since {{{ w^2 = cos (4*pi/3) + i sin (4*pi/3) }}}
              = {{{ cos (pi+ pi/3) + i sin (pi+pi/3)}}}
              = {{{ -cos (pi/3) - i sin (pi/3)}}}
         and {{{ w = cos (2*pi/3) + i sin (2*pi/3) }}}
             = {{{ w = cos (pi- pi/3) + i sin (pi-pi/3) }}}
             = {{{ w = -cos (pi/3) + i sin (pi/3) }}}

    When n =2, {{{ 1+ w^2 + w^4 = 1 + w^2 + w }}}
    [Note: {{{ w^4 = w}}}]    
    = {{{ 1+ w + w^2 = 1 -cos (pi/3) -cos (pi/3) }}}
       = {{{ 1-1/2 -1/2  = 0 }}}
     {or use ew is the primitive root of {{{ 1+x+x^2 = 0}}} directly]
  
   Similarly, when n =4, {{{ 1+ w^4 + w^8 = 1 + w + w^2 = 0}}}
    [Note: {{{ w^8 = w^2}}}]  


 Kenny