Solve equation with complex numbers using a helper equation z 2 = 5 − 12Let Let z = x + y i ( x + y i ) 2 = 5 − 12 i x 2 − y 2 + 2 i x y = 5 − 12 iThis means that we get the following system of equations: { x 2 − y 2 = 5 2 x y = 12 | z 2 | = 5 2 − 12 2 = 13If If z = x + y i then | z | 2 = 13 ⇔ x 2 + y 2 = 13 then If z = x + y i then | z | 2 = 13 ⇔ x 2 + y 2 = 13

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Solve equation with complex numbers using a helper equation      z    2    =  5  −  12Let   Let     z  =  x  +  y  i  (  x  +  y  i      )    2    =  5  −  12  i      x    2    −      y    2    +  2  i  x  y  =  5  −  12  iThis means that we get the following system of equations:      {                                        x            2                    −                      y            2                    =          5                                      2          x          y          =          12                              |        z    2        |    =            5      2        −          12      2        =  13If   If     z  =  x  +  y  i     then         |    z            |        2    =  13  ⇔      x    2    +      y    2    =  13 then   If     z  =  x  +  y  i     then         |    z            |        2    =  13  ⇔      x    2    +      y    2    =  13

Jackson Trevino · Accepted Answer

You can solve the system      {                                        x            2                    +                      y            2                    =          13                                                  x            2                    −                      y            2                    =          5                        for       x    2   and       y    2  . This just gives you a very few possibilities for   x and   y, which must also satisfy   x  y  =  6.

kituoti126 · Accepted Answer

(      x    2    +      y    2        )    2    =  (      x    2    −      y    2        )    2    +  (  2  x  y      )    2    =      5    2    +      12    2    =      13    2      ⟹        x    2    +      y    2    =  13We already have       x    2    −      y    2    =  5As   x  y  =  6  &amp;gt;  0  ;  x  ,  y must be of same sign

Solve equation with complex numbers using a helper equation, z^2=5−12, Let z=x+yi, (x+yi)^2=5−12i, x^2−y^2+2ixy=5−12i

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