Proof of commutativity for complex numbers help

I just started Linear Algebra and my algebra is a little rusty. However, was going over the proof of associativity for complex number and while I was able to do it, I got stuck on the way they do it in the book.

Let $$\alpha=x_1+y_1i\\ \beta=x_2+y_2i\\ \lambda=x_3+y_3i\\$$

where $$x_1,x_2,x_3$$ and $$y_1,y_2,y_3$$ are real numbers. Then \begin{align} (\alpha\beta)\lambda &= ((x_1x_2−y_1y_2)+(x_1y_2+y_1x_2)i)(x_3+y_3i)\\&=((x_1x_2−y_1y_2)x_3−(x_1y_2+y_1x_2)y_3)+((x_1x_2−y_1y_2)x_3+(x_1y_2+y_1x_2)y_3)i. \end{align}

The part that I am struggling with here is: \begin{align} &=((x_1x_2−y_1y_2)x_3−(x_1y_2+y_1x_2)y_3)+((x_1x_2−y_1y_2)x_3+(x_1y_2+y_1x_2)y_3)i. \end{align}

Basically I don't understand how we go here from the previous line. Maybe I have been staring at this for a while.

• With $x1$ you probably mean $x_1$ (written x_1) etc. Please correct this and also delete unnecessary repetitions to improve the readability of the post. – Claudius Apr 28 '17 at 17:35
$$(x+iy)(z+iw)=xz+i(wx+yz)-yw$$ Quite similarly $$(z+iw)(x+iy)=zx+i(zy+xw)-wy$$ since each of the products are real and thus commute, we have equality of real ad imaginary parts in the two expressions.