CR Equations using Polar Form.

Question

I have a question to check whether following function is analytic or not using CR Equations. The question is

$$ f(z) = {1 \over(z-z^5)}$$

I just don't know how to start and separate them into real and imaginary parts. That seems kind of an easy question but i really am new to complex planes and so there are so many questions of this type causing problem to me. Thanks for help in advance

Answer 1

Notice $z = re^{i \theta}$. Let me take a different approach than my comment, $$ \frac{1}{z^5-z} = \frac{1}{z}\frac{1}{z^4-1} $$ So, to put $1/z$ into Cartesian form multiply by $\bar{z}/\bar{z}$ and multiply by $\frac{\bar{z}^4-1}{\bar{z}^4-1}$. $$ \frac{1}{z^5-z} = \frac{\bar{z}}{z\bar{z}}\frac{\bar{z}^4-1}{(z^4-1)(\bar{z}^4-1)} = \frac{\bar{z}^5-\bar{z}}{z\bar{z}((z\bar{z})^4-z^4-\bar{z}^4+1)} $$ Ok, so $z = re^{i \theta}$ and $\bar{z} = re^{-i \theta}$ hence $z\bar{z}=r^2$ and $z^4 = r^4e^{4i \theta}=r^4(\cos 4 \theta+i \sin 4\theta)$ likewise $\bar{z}^4 = r^4e^{-4i \theta}=r^4(\cos 4 \theta-i \sin 4\theta)$ so we see $\bar{z}^4+z^4 =2r^4\cos \theta$. Thus, \begin{align} \frac{1}{z^5-z} &= \frac{r^5(\cos 5 \theta-i\sin 5\theta)-r(\cos \theta-i\sin \theta)}{r^2(r^8-2r^4\cos 4\theta+1)} \\ &= \underbrace{\frac{r^5\cos 5 \theta-r\cos \theta}{r^2(r^8-2r^4\cos 4\theta+1)}}_{U(r, \theta)} + i\underbrace{\left(\frac{-r^5\sin 5\theta+r\sin \theta}{r^2(r^8-2r^4\cos 4\theta+1)}\right)}_{V(r, \theta)} \end{align} There might be some easier way through the algebra, but, as you can see, polar CR-equations are probably not a good choice of analytical technique on this problem. I might ask it, but, just to build skill in trig and algebra manipulation.

Answer 2

By using the notation $z$ as a variable I presume by analytical you mean holomorphic. The condition for a function $f$ to be holomorphic in a point $z_0\in \Bbb{C}$ is that the limit $\lim_{\epsilon \to 0 } \frac{f(z_0+\epsilon)-f(z_0)}{\epsilon}$ exists. This implies not only that $\frac{\partial f_r}{\partial x}$, $\frac{\partial f_i}{\partial x}$,$\frac{\partial f_r}{\partial y}$ and $\frac{\partial f_i}{\partial y}$ exist but also satisfy the Cauchy-Riemann conditions. These roughly tell us that the linear approximation of the function in a certain point is the multiplication with a complex number $a+\iota b$ and so in the complex plane takes the form $\bigl( \begin{smallmatrix} a & -b \\ b & a\end{smallmatrix} \bigr)$ i.e. a rotation and a scaling, and not a skew transformation. In practice it is enough when the function is expressed in $z$ only, without $\bar{z}$. This is the case of your function and $f'(z)=(5 z^4-1)/(z^2 (z^4-1)^2)$ which shows that $f$ is holomorphic everywhere except at the origin. Functions like $z+\bar{z}$ and $z/\bar{z}$ are not holomorphic.