Functions of complex variable and its differentiation

Question

w=f(z) (complex function) w=u +iv (u, v are funtions of x and y) How to find dw/dz in polar form? Can someone please explain me the chain rule expansion involved in this The derivative should be equal to partial differential of w in terms of r, theta and x

and z=re^(itheta)

Answer 1

The total differential of $w=f(z)=:f(x,y)=u(x,y)+i(v,x)$ is \begin{align*} df &= \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y}dy\\ &=\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)dx + \left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)dy\tag{1} \end{align*}

With $x=r\cos \theta$ and $y=r\sin\theta$ we obtain \begin{align*} \frac{\partial u}{\partial x} &=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial\theta}{\partial x} =\frac{\partial u}{\partial r}\frac{x}{\sqrt{x^2+y^2}}-\frac{\partial u}{\partial \theta}\frac{y}{x^2+y^2}\\ &=\cos\theta\frac{\partial u}{\partial r}-\frac{1}{r}\sin\theta\frac{\partial u}{\partial \theta}\tag{2}\\ \frac{\partial u}{\partial y} &=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial \theta}\frac{\partial\theta}{\partial y} =\frac{\partial u}{\partial r}\frac{y}{\sqrt{x^2+y^2}}+\frac{\partial u}{\partial \theta}\frac{x}{x^2+y^2}\\ &=\sin\theta\frac{\partial u}{\partial r}+\frac{1}{r}\cos\theta\frac{\partial u}{\partial \theta}\tag{3} \end{align*} and similarly with $v(x,y)$ instead of $u(x,y)$.

We obtain with $x=x(r,\theta)=r\cos \theta$ and $y=y(r,\theta)=r\sin\theta$ \begin{align*} dx&=\frac{\partial x}{\partial r} dr+\frac{\partial x}{\partial \theta}d\theta =\cos\theta\,dr-r\sin\theta\,d\theta\tag{4}\\ dy&=\frac{\partial y}{\partial r} dr+\frac{\partial y}{\partial \theta}d\theta =sin\theta\,dr+r\cos\theta\,d\theta\tag{5}\\ \end{align*}

Putting (2) - (5) into (1) we obtain \begin{align*} \color{blue}{df}&\color{blue}{=\left(\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}\right)dx + \left(\frac{\partial u}{\partial y}+i\frac{\partial v}{\partial y}\right)dy}\\ &=\left(\cos\theta\frac{\partial u}{\partial r}-\frac{1}{r}\sin\theta \frac{\partial u}{\partial \theta}\right.\\ &\qquad\left.+i\left(\cos\theta\frac{\partial v}{\partial r}-\frac{1}{r}\sin \theta\frac{\partial v}{\partial \theta}\right)\right)\left(\cos\theta\,dr-r\sin\theta\,d\theta\right)\\ &\quad+\left(\sin\theta\frac{\partial u}{\partial r}+\frac{1}{r}\cos\theta \frac{\partial u}{\partial \theta}\right.\\ &\qquad\left.+i\left(\sin\theta\frac{\partial v}{\partial r}+\frac{1}{r}\cos \theta\frac{\partial v}{\partial \theta}\right)\right)\left(\sin\theta\,dr+r\cos\theta\,d\theta\right)\\ &=1\cdot\frac{\partial u}{\partial r}dr+0\cdot\frac{\partial u}{\partial \theta}dr +0\cdot\frac{\partial u}{\partial r}d\theta+1\cdot\frac{\partial u}{\partial \theta}d\theta\\ &\quad+i\left(1\cdot\frac{\partial v}{\partial r}dr+0\cdot\frac{\partial v}{\partial \theta}dr +0\cdot\frac{\partial v}{\partial r}d\theta+1\cdot\frac{\partial v}{\partial \theta}d\theta\right)\tag{6}\\ &\,\,\color{blue}{=\left(\frac{\partial u}{\partial r}+i\frac{\partial v}{\partial r}\right)dr+\left(\frac{\partial u}{\partial \theta}+i\frac{\partial v}{\partial \theta}\right)d\theta} \end{align*} In (6) we use $\sin^2 \theta+\cos^2 \theta=1$ (factor $1$) and observe that terms cancel (factor $0$).