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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 of the function $z = z(r,\theta)$, so $\displaystyle dz =\frac{ \partial z }{\partial r}\,dr + \frac{\partial z}{\partial \theta} \,d\theta$. – t.b. Jul 8 '11 at 22:34
You can think of $z$ as being a function of the parameters $r$ and $\theta$. Now take the total derivative: – tomcuchta Jul 8 '11 at 22:35
$ d $ doesn't stand for derivative, it stands for differential. – anon Jul 8 '11 at 22:35
Wow, talk about a flurry. :-) (And interesting that everyone chose to answer in comments rather than as an answer...) – Steven Stadnicki Jul 8 '11 at 22:37
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. – Américo Tavares Jul 8 '11 at 22:59

The comments so far leave unclear what we are talking about when we write $dz$, $dr$, etc. So let me elaborate on this point a bit.

To any function $f:\ \Omega\to{\mathbb C}$ on some open set $\Omega\subset{\mathbb R}^2$ is associated its differential $df$. The differential $df$ measures how much the value of the function changes when you move from a point $p\in\Omega$ to a nearby point $p+h$. Now this change is in first approximation a linear function of the displacement vector $h$. This linear map from the tangent space at $p$ to ${\mathbb C}$ (whence a functional) is by definition the differential of $f$ at $p$. Written as a formula this is to say that $$f(p+h)-f(p)\ =\ df(p).h + o(|h|)\qquad (h\to 0)\ .$$ Now $z(\cdot)$, $r(\cdot)$ and $\theta(\cdot)$ are functions on ${\mathbb C}$ in their own right, and so they have differentials $dz$, $dr$, $d\theta$. The formula $$(*) \qquad dz=e^{i\theta}dr + ir e^{i\theta}d\theta$$ says that at each point $p=r e^{i\theta}\in{\mathbb C}$ the three functionals $dz(p)$, $dr(p)$ and $d\theta(p)$ are related in a particular way.

Now therewith the formula $(*)$ is not proven yet. Suffice it to say that the "rules of calculus" about handling differentials can indeed be justified with not much effort, and the formula is true . . .

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You are assuming that $r$ and $\theta$ are both functions of some variable $t$, and then using garden variety "sum rule" from single variable calculus.

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