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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.

Thank you in advance!

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    $\begingroup$ 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. $\endgroup$ – Claudius Apr 28 '17 at 17:35
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    $\begingroup$ Also, you are checking associativity and not commutativity. $\endgroup$ – Claudius Apr 28 '17 at 17:36
  • $\begingroup$ Do you know that the property holds for the real numbers? How do you define complex multiplication? $\endgroup$ – Stella Biderman Apr 28 '17 at 18:56
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$$ (x+iy)(z+iw)=xz+i(wx+yz)-yw $$ 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.

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  • $\begingroup$ op is trying to check associativity, not commutativity $\endgroup$ – Matthew Towers Apr 28 '17 at 19:26
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    $\begingroup$ Do we know what the question is asking? $\endgroup$ – operatorerror Apr 28 '17 at 19:39

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