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Prove that if we write $z = re^{i\theta}$, then $d$ for derivative, $$dz=e^{i\theta}\,dr + ire^{i\theta}\,d{\theta}$$

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    Well, it's just the [total derivative](http://en.wikipedia.org/wiki/Total_derivative) of the function $z = z(r,\theta)$, so $\displaystyle dz =\frac{ \partial z }{\partial r}\,dr + \frac{\partial z}{\partial \theta} \,d\theta$.2011-07-08
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    You can think of $z$ as being a function of the parameters $r$ and $\theta$. Now take the total derivative: http://en.wikipedia.org/wiki/Total_derivative2011-07-08
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    Product rule, Chain Rule.2011-07-08
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    $ d $ doesn't stand for derivative, it stands for differential.2011-07-08
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    This identity is just the product formula for the derivative/differential, treating $r$ and $\theta$ as unknowns - that is, that $d(ab) = (da)b+a(db)$, here with $a=r$ and $b=e^{i\theta}$ (along with a little bit of the chain rule to compute $db$).2011-07-08
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    Wow, talk about a flurry. :-) (And interesting that everyone chose to answer in comments rather than as an answer...)2011-07-08
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    For a function $y = f(x_1,\dots,x_n)$ of several variables the *total differential* is $dy = \frac{\partial y}{\partial x_1} dx_1 + \cdots + \frac{\partial y}{\partial x_n} dx_n$ [Wikipedia](http://en.wikipedia.org/wiki/Differential_%28calculus%29#Differentials_in_several_variables).2011-07-08

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