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Analyticity & Cauchy–Riemann

Useful preparation: Integrals & the Fundamental Theorem

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Meaning & notation

A complex derivative describes change from every direction using one complex number. Its real linear approximation must therefore act like complex multiplication.

Complex differentiability: For a function defined near $\displaystyle z_{0}$ , the derivative is the limit of $\displaystyle \frac{f\, \left(z_{0} + h\right) - f\, \left(z_{0}\right)}{h}$ as the nonzero complex increment h tends to zero. The limit must be independent of the approach.
Holomorphic and analytic: A function is holomorphic on an open set if it is complex differentiable at every point there. In one complex variable this is equivalent to locally having a convergent power series, called analyticity.
Cauchy–Riemann equations: Write $\displaystyle z = x$ +iy and $\displaystyle f = u$ +iv with real-valued u,v. The equations $\displaystyle u_{x} = v$ ᵧ and uᵧ= $\displaystyle -v_{x}$ connect their partial derivatives; subscripts specify the real coordinate differentiated.

Conditions before calculation

Complex differentiability at a point implies the Cauchy–Riemann equations there. Their satisfaction alone at one point is not a sufficient differentiability test.
A sufficient pointwise test is that u and v have continuous first partial derivatives in a neighborhood and satisfy the equations at the point. If they satisfy them throughout an open set, f is holomorphic there.
Differentiability at one isolated point does not imply holomorphicity on any neighborhood. When using quotient or logarithm formulas, exclude zeros or specify an appropriate branch and domain.

Why the method works

1. Compare two necessary limits

Approach with a real increment and then a purely imaginary increment. If the complex derivative exists, both resulting expressions must agree, giving the equations.

f'(z_0)=u_x+iv_x=v_y-iu_y

2. Use real differentiability for sufficiency

Continuous partials provide a real linear approximation. Its error divided by $\displaystyle \left|h\right|$ tends to zero, denoted $\displaystyle o\, \left(\left|h\right|\right)$ . The equations turn the linear part into multiplication by $\displaystyle u_{x}$ +ivₓ.

f(z_0+h)-f(z_0)=(u_x+iv_x)h+o(|h|)

3. Check the scope of the conclusion

After computing a derivative, state where it exists. Passing the test only at one point supports only a pointwise conclusion, not analyticity of a whole region.

Worked examples

EXAMPLE 1

Verify that $\displaystyle f\, \left(z\right) = z^{2} + \left(2 - i\right)\, z$ is holomorphic and compute $\displaystyle f'\, \left(1 - i\right)$ .

Expand using $\displaystyle z = x$ +iy to identify the real and imaginary parts.
$u=x^2-y^2+2x+y,\qquad v=2xy+2y-x$
The partials are continuous polynomials and satisfy both equations everywhere.
$u_x=v_y=2x+2,\qquad u_y=-2y+1=-v_x$
Use $\displaystyle u_{x}$ +ivₓ for the derivative and substitute the requested point.
$f'(z)=2z+2-i,\qquad f'(1-i)=4-3i$

Conclusion: The polynomial is holomorphic on the entire plane, and the derivative is $\displaystyle 4 - 3\, i$ at $\displaystyle 1 - i$ .

EXAMPLE 2

Determine where $\displaystyle g\, \left(z\right) = \left|z - 2\right|^{2}$ is complex differentiable.

Here $\displaystyle u = \left(x - 2\right)^{2} + y^{2}$ and $\displaystyle v = 0$ . The equations force $\displaystyle x = 2$ and $\displaystyle y = 0$ , so no other point can work.
At $\displaystyle z = 2$ , evaluate the difference quotient directly.
$\frac{g(2+h)-g(2)}h=\frac{|h|^2}h=\overline h$
Its modulus equals $\displaystyle \left|h\right|$ and tends to zero along every approach, proving $\displaystyle g'\, \left(2\right) = 0$ .

Conclusion: It is differentiable only at 2 and is holomorphic on no nonempty open set.

Pitfalls & counterexamples

Tempting claim: Cauchy–Riemann equations at one point always prove differentiability.

Correction: A regularity assumption is needed for this converse.

Test the claim: Set $\displaystyle u = \frac{xy}{\sqrt{x^{2} + y^{2}}}$ away from the origin, $\displaystyle u\, \left(0, 0\right) = 0$ , and $\displaystyle v = 0$ . All partials at zero vanish, but the quotient is zero along real increments and $\displaystyle \frac{1}{\sqrt{2}\, \left(1 + i\right)}$ along $\displaystyle h = t\, \left(1 + i\right)$ , $\displaystyle t > 0$ .

Check your understanding

Is the conjugation map $\displaystyle f\, \left(z\right) = z$ ̄ complex differentiable anywhere?

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Compare your reasoning

No.

Its difference quotient is 1 for nonzero real increments and −1 for purely imaginary increments.

Connect this idea

The equations connect complex differentiation with real Jacobian matrices. Holomorphicity on regions is the foundation for Cauchy’s theorem, local power series, and residue calculations.

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Reference reading

Use these texts to study the underlying definitions and theorems in greater depth. The explanations and examples above are written for Numerive.

MIT OCW 18.04, Topic 2: Analytic functions