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Doing an exercise on complex analysis where it began by asking me to solve some equations for $u_x$ and $u_y$ I got stuck and looked up the answer.

$$u_x\cos\theta+u_y\sin\theta=u_r,$$ $$-u_xr\sin\theta+u_yr\cos\theta=u_\theta.$$ Solving these simultaneous linear equations for $u_x$ and $u_y$, we find that $$u_x=u_r\cos\theta-u_\theta\frac{\sin\theta}{r}\quad\text{ and }\quad u_y=u_r\sin\theta+u_\theta\frac{\cos\theta}{r}.$$

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The answer says 'solving these simultaneous linear equations'...but surely these are not linear equations when they use cos and sin?

And how do you solve these equations anyway, i can see you can divide across the bottom equation but I can see what to do after that.

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    $\begingroup$ $\theta$, $u_\theta$, $u_r$, and $r$ are treated as constants. Note, the solution is different for different $\theta$, $u_\theta$, and $u_r$, $r$. $\endgroup$ – David Mitra Feb 11 '12 at 18:43
  • $\begingroup$ They're not "linear in $\theta$" since $\sin$ and $\cos$ are involved. They're linear in $u_x$ and $u_y$. $\endgroup$ – Michael Hardy Feb 11 '12 at 20:28
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The equations are linear in $u_x$ and $u_y$—that is, if $\theta$ is a constant, $\sin\theta$ and $\cos\theta$ are constants, so they're just like having numbers there.

If you multiply the first equation by $\sin\theta$ and the second equation by $\frac{\cos\theta}{r}$, you'll have $$u_x\cos\theta\sin\theta+u_y\sin^2\theta=u_r\sin\theta$$ $$-u_x\cos\theta\sin\theta+u_y\cos^2\theta=u_\theta\frac{\cos\theta}{r}$$ Adding these two equations gives $$u_y(\sin^2\theta+\cos^2\theta)=u_r\sin\theta+u_\theta\frac{\cos\theta}{r}$$ and since $\sin^2\theta+\cos^2\theta=1$, $$u_y=u_r\sin\theta+u_\theta\frac{\cos\theta}{r}.$$

You can find $u_x$ from this $u_y$, or by applying a similar technique to the original equations.

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