How do I show that $w = \frac{1}{(z-a)^n}$ is an analytic function using the Cauchy Riemann equations,where z = x + iy?

Question

I know I need to use the Cauchy Riemann equations, to test for the analyticity, but how do I break the function $w = \frac{1}{(z-a)^n}$ into real and imaginary parts?

Answer 1

With

$z = x + iy, \tag 1$

$a = \sigma + i\omega, \tag 2$

$z - a = (x - \sigma) + i(y - \omega); \tag 3$

$\dfrac{1}{z - a} = \dfrac{1}{(x - \sigma) + i(y - \omega)} = \dfrac{(x - \sigma) - i(y - \omega)}{(x - \sigma)^2 + (y - \omega)^2}$ $= \dfrac{x - \sigma}{(x - \sigma)^2 + (y - \omega)^2} - i\dfrac{y - \omega}{(x - \sigma)^2 + (y - \omega)^2} = u(x, y) + iv(x, y); \tag 4$

it follows that

$u(x, y) = \dfrac{x - \sigma}{(x - \sigma)^2 + (y - \omega)^2} \tag 5$

and

$v(x, y) = - \dfrac{y - \omega}{(x - \sigma)^2 + (y - \omega)^2} \tag 6$

are the real and imaginary parts of $(z - a)^{-1}$, respectively. At this point it is a relatively easy matter to apply the Cauchy-Riemann equations to $u(x, y)$ and $v(x, y)$ directly, though the necessary algebra is a bit tedious:

writing

$u(x, y) = (x - \sigma)((x - \sigma)^2 + (y - \omega)^2)^{-1}, \tag 7$

$v(x, y) = - (y - \omega)((x - \sigma)^2 + (y - \omega)^2)^{-1}, \tag 8$

we have

$u_x(x, y) = ((x - \sigma)^2 + (y - \omega)^2)^{-1} -2(x - \sigma)^2((x - \sigma)^2 + (y - \omega)^2)^{-2}$ $ = \dfrac{1}{(x - \sigma)^2 + (y - \omega)^2} - \dfrac{2(x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$ $= \dfrac{(x - \sigma)^2 + (y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2} - \dfrac{2(x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$ $= \dfrac{(y - \omega)^2 - (x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}; \tag 9$

we also have

$v_y(x, y) = -((x - \sigma)^2 + (y - \omega)^2)^{-1} + 2(y - \omega)^2((x - \sigma)^2 + (y - \omega)^2)^{-2}$ $ = -\dfrac{1}{(x - \sigma)^2 + (y - \omega)^2} + \dfrac{2(y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$ $= -\dfrac{(x - \sigma)^2 + (y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2} + \dfrac{2(y - \omega)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}$ $= \dfrac{(y - \omega)^2 - (x - \sigma)^2}{((x - \sigma)^2 + (y - \omega)^2)^2}; \tag{10}$

comparing (9) and (10) we see that

$u_x(x, y) = v_y(x, y); \tag{11}$

a similar calculatinon validates

$u_y(x, y) = -v_x(x, y), \tag{12}$

and thus, with of course the proviso that $z \ne a$,

$\dfrac{1}{z - a} = u(x, y) + iv(x, y) \tag{13}$

satisfies the Cauchy-Riemann equations, and hence is holomorphic.

The preceding calculations may be simplified somewhat if we express $(z - a)^{-1}$ in term of a polar coordinate system centered at $a$; that is, we write

$z = a + re^{i\theta}, \tag{14}$

whence

$\dfrac{1}{z - a} = (z - a)^{-1}$ $= (re^{i\theta})^{-1} = r^{-1}e^{-i\theta} = \dfrac{\cos \theta}{r} - i\dfrac{\sin \theta}{r} = u(r, \theta) + iv(r, \theta); \tag{15}$

the Cauchy-Riemann equations in such a polar coordinate system are given by

$u_r = \dfrac{1}{r}v_\theta, \tag{16}$

$v_r = -\dfrac{1}{r}u_\theta; \tag{17}$

now a relatively simple calculation shows that, with $u(r, \theta)$ and $v(r, \theta)$ as in (15),

$\left ( \dfrac{\cos \theta}{r} \right)_r = -\dfrac{\cos \theta}{r^2} = \dfrac{1}{r} \left ( -\dfrac{\sin \theta}{r} \right)_\theta;$ $\left ( -\dfrac{\sin \theta}{r} \right)_r = \dfrac{\sin \theta}{r^2} = -\dfrac{1}{r} \left ( \dfrac{\cos \theta}{r} \right)_\theta, \tag{18}$

again establishing that $(z - a)^{-1}$ is holomorphic away from $z = a$, though with much less effort.

Once we have that $(z - 1)^{-1}$ is a holomorphic function, we may simply invoke the well-known fact that the product of holomorphic functions is again holomorphic to affirm that

$\dfrac{1}{(z - a)^2} = \dfrac{1}{z - a} \dfrac{1}{z - a} \tag{19}$

is holomorphic, and then use a simple inductive argument, assuming that $(z - a)^{-k}$ is holomorphic for $k \in \Bbb N$, to see that

$\dfrac{1}{(z - a)^{k + 1}} = \dfrac{1}{(z - a)^k}\dfrac{1}{z - a} \tag{20}$

is also holomrphic; we conclude that $(z - a)^{-n}$ is holomorphic for all $n \in \Bbb N$. $OE\Delta$.

Answer 2

For $z_0≠a$ you have a quotient of analytic functions with non-vanishing denominator in a neighborhood of $z_0$. Hence it is analytic at $z_0$.(It is not hard to prove this for a general quotient $f/g$ by definition of the derivative)