Neutral Element on $(\mathbb{C},*,+)$
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?
$\endgroup$ 42 Answers
$\begingroup$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 $e_y=0$
$\endgroup$ 5 $\begingroup$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 $(1,0)$
$\endgroup$
