Respuesta :
Given:
There are given the complex equation:
[tex]2(1-i)z^2+(1+i)z+3(1-i)=0[/tex]Explanation:
To find the value, we need to put the standard value of z into the above-given complex function:
[tex]\begin{gathered} 2(1-\imaginaryI)z^2+(1+\imaginaryI)z+3(1-\imaginaryI)=0 \\ 2(1-\imaginaryI)(x+yi)^2+(1+\imaginaryI)(x+yi)+3(1-\imaginaryI)=0 \end{gathered}[/tex]Then,
[tex]\begin{gathered} 2(1-\imaginaryI)(x+y\imaginaryI)^2+(1+\imaginaryI)(x+y\imaginaryI)+3(1-\imaginaryI)=0 \\ (2x^2-2y^2+4xy+x-y+3)+i(-3+x+y-2x^2+2y^2+4xy)=0 \end{gathered}[/tex]Then,
[tex](2x^2-2y^2+4xy+x-y+3)+\imaginaryI(-3+x+y-2x^2+2y^2+4xy)=0+0i[/tex]So,
[tex]\begin{gathered} (2x^2-2y^2+4xy+x-y+3)+\imaginaryI(-3+x+y-2x^2+2y^2+4xy)=0+0i \\ 2x^2-2y^2+4xy+x-y+3=0 \\ -3+x+y-2x^2+2y^2+4xy=0 \end{gathered}[/tex]Then,
[tex]\begin{gathered} x=0,y=-\frac{3}{2} \\ x=0,y=1 \end{gathered}[/tex]Final answer:
Hence, the value of the given complex function is shown below:;
[tex]\begin{gathered} z=-\frac{3}{2}i \\ z=i \end{gathered}[/tex]