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Evaluate and simplify the product

$\begin{bmatrix} r\cos(\alpha) & -r\sin(\alpha) \\ r\sin(\alpha) & r\cos(\alpha)\\ \end{bmatrix}$ $\begin{bmatrix} s\cos(\beta) & -s\sin(\beta) \\ s\sin(\beta) & s\cos(\beta)\\ \end{bmatrix}$

where $\alpha$, $\beta$ are any real numbers and $r$, $s$ are nonnegative real numbers. Relate your answer to multiplication of complex numbers. Is there any connection with addition of complex numbers?

I was able to multiply through and I will post my answer below, but I can't see the connection to complex numbers. Could someone please explain? Thanks.

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  • $\begingroup$ tip for future formatting: trig functions format much more nicely if you add a backslash immediately preceding the function. Compare, e.g., r\cos(\alpha) which gives you $r\cos(\alpha)$ vs. rcos(\alpha) which gives $rcos(\alpha)$ $\endgroup$
    – amWhy
    May 19, 2013 at 1:46

2 Answers 2

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Using the fact that $e^{i\theta}=\cos\theta+i\sin\theta$, what do you get when you multiply $$re^{i\alpha}\cdot se^{i\beta}?$$Given the matrix $$\left[\begin{array}{rr} a & -b \\ b & a \end{array}\right]$$ what complex number would you associate it with?

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    $\begingroup$ would it be $a-bi$? $\endgroup$
    – Joe S
    May 19, 2013 at 2:03
  • $\begingroup$ @JoeS Although that would still work, the convention is to associate it with $a+bi$, not the complex conjugate. $\endgroup$
    – Joseph G.
    May 19, 2013 at 2:04
  • $\begingroup$ ok, thanks for your help $\endgroup$
    – Joe S
    May 19, 2013 at 2:09
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$\begin{bmatrix} rs\cos(\alpha+\beta) & -rs\sin(\alpha+\beta) \\ rs\sin(\alpha+\beta) & rs\cos(\alpha+\beta)\\ \end{bmatrix}$

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