Question 1181502


roots of 

{{{z^2 - z(1 + i) + 5i = 0}}}......... substitute {{{z=x+yi}}}


{{{(x+yi)^2 - (x+yi)(1 + i) + 5i = 0}}}...expand


{{{x^2+2xy*i+(yi)^2 - (x+x*i+yi+y*i^2)+ 5i = 0}}}


{{{x^2+2xy*i+(yi)^2 - (x+x*i+yi-y)+ 5i = 0}}}


{{{x^2+2xy*i+(y^2i^2) - x-x*i-yi+y+ 5i = 0}}}


{{{x^2+2xy*i-y^2 - x-x*i-yi+y+ 5i = 0}}}...........combine real parts and imaginary parts


{{{(x^2-y^2 -x+y)+2xy*i-x*i-yi+ 5i=0}}}


{{{(x^2-y^2 -x+y)-i(2xy-x-y+5)=0}}}


Complex numbers can be equal only if their real and imaginary parts are equal.

{{{x^2-y^2 -x+y=0}}}.....eq.1
{{{2xy-x-y+5=0}}}.....eq.2
-------------------solve this system 

{{{2xy-x=y-5}}}....eq.2, solve for {{{x}}}

{{{(2y-1)x=y-5}}}

{{{x=(y-5)/(2y-1)}}}

substitute in eq.1

{{{((y-5)/(2y-1))^2-y^2 -(y-5)/(2y-1)+y=0}}}

{{{((y - 2) (y + 1))/(2 y - 1) = 0}}}} .........zeros are in numerator

{{{(y - 2) (y + 1)=0}}}

{{{y=2 }}}or {{{y=-1}}}

then

{{{x=(2-5)/(2*2-1)=-3/3=-1}}}

{{{x=(-1-5)/(2(-1)-1)=-6/-3=2}}}

solutions:


{{{x=2}}}, {{{y=-1}}}
or
{{{x=-1}}}, {{{y=2}}}

Substitute back  {{{z=x+yi}}}, and roots are

{{{z[1]=2-i}}}
{{{z[2]=-1+2i}}}