Neutral Element on $(\mathbb{C},*,+)$

Question

I want to show the neutral Element on $$(\mathbb{C},*,+)$$

Let the complex Number be defined as: $$(x,y)$$ Multiplication of complex numbers is defined as: $$(x_{1}, y_{1}) * (x_{2}, y_{2}):= (x_{1}x_{2}-y_{1}y_{2}, x_{1}y_{2}+x_{2}y_{1})$$ By definition the neutral element is: $$(e_{x}, e_{y}) *(x,y) = (x,y)$$ $$(x*e_{x}-y*e_{y}, x*e_{y}+y*e_{x})$$ This gives an equation system: $$x = x*e_{x} -y*e_{y}$$ $$y = x*e_{y} +y*e_{x}$$

Is this the correct approach to finding the neutral Element?

If yes, what is the correct method for solving the equation system?

Answer 1

The answer is $$\mathrm{I}\quad x*e_x-y*e_y = x \quad|*x$$ $$\mathrm{II}\quad x*e_y+y*e_x = x \quad|*y$$ $$\mathrm{I+II} \quad x^2*e_x+y^2*e_x = x^2 + y^2$$ $$\Leftrightarrow \quad e_x * (x^2 + y^2) = x^2 + y^2 \quad | :(x^2 + y^2 )$$ $$\Leftrightarrow \quad e_x = 1$$ Putting the solution $e_x = 1 $ into $\mathrm{I}$ $$x*1-y *e_y = x \quad |-x$$ $$\Leftrightarrow \quad-y*e_y=0 \quad|:(-y)$$ $$\Leftrightarrow \quad e_y=0 $$ Thus the neutral element is $(1,0)$

Answer 2

Your approach is not (quite) correct because you should have $x=xe_x-ye_y$ and $y=xe_y+ye_x$ (just normal multiplication instead of your "*" operation). Also, the equations should hold for any x and y in $\mathbb{C}$ and this implies that $e_x=1$ and $e_y=0$