Show 1/z⁴ is analytic except z = 0

Analytic function

f(z) is analytic at a point if it is differentiable at that point and at all points in some neighbourhood of it.

Method 1: C–R equations in polar form

Let z = re^{iθ}. Then f(z) = z⁻⁴ = r⁻⁴ e^{−4iθ} = r⁻⁴ cos 4θ − i r⁻⁴ sin 4θ.

u = r⁻⁴ cos 4θ, v = −r⁻⁴ sin 4θ.

u_r = −4 r⁻⁵ cos 4θ        u_θ = −4 r⁻⁴ sin 4θ
v_r =  4 r⁻⁵ sin 4θ        v_θ = −4 r⁻⁴ cos 4θ

Check the polar C–R equations (valid for r ≠ 0):

• u_r = −4r⁻⁵ cos 4θ and (1/r) v_θ = −4r⁻⁵ cos 4θ. Equal ✓
• v_r = 4r⁻⁵ sin 4θ and −(1/r) u_θ = 4r⁻⁵ sin 4θ. Equal ✓

All partial derivatives are continuous for r ≠ 0, so the C–R equations are satisfied and sufficient conditions hold. Hence f is analytic at every z ≠ 0.

Method 2: direct differentiation

For z ≠ 0,

f′(z) = lim_{h→0} [ (z + h)⁻⁴ − z⁻⁴ ] / h = lim [ z⁴ − (z + h)⁴ ] / [ h z⁴ (z + h)⁴ ] = −4z³/(z⁴ · z⁴) = −4/z⁵.

The limit exists at every z ≠ 0.

At z = 0, f is not even defined, so it is not differentiable there.

Conclusion: f(z) = 1/z⁴ is analytic everywhere except at z = 0, and f′(z) = −4/z⁵.