Question

To resolve 2(1-i) z² + (1+i) z + 3 (1 - i) = 0

Answer 1

Given:

There are given the complex equation:

`2(1-i)z^2+(1+i)z+3(1-i)=0`

Step-by-step explanation:

To find the value, we need to put the standard value of z into the above-given complex function:

`\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}`

Then,

`\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}`

Then,

`(2x^2-2y^2+4xy+x-y+3)+\imaginaryI(-3+x+y-2x^2+2y^2+4xy)=0+0i`

So,

`\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}`

Then,

`\begin{gathered} x=0,y=-(3)/(2) \\ x=0,y=1 \end{gathered}`

Final answer:

Hence, the value of the given complex function is shown below:;

`\begin{gathered} z=-(3)/(2)i \\ z=i \end{gathered}`