Complex Analysis 7 | Cauchy-Riemann Equations Examples [dark version]

TL;DR

Write any complex function as f(z)=u(x,y)+i v(x,y) by substituting z=x+iy and separating real and imaginary parts.

Briefing Cornell Notes

Briefing

Cauchy–Riemann equations turn the question “Is a complex function holomorphic?” into checking two partial differential equations for its real and imaginary parts. Once a complex function f=u+iv is written in terms of u(x,y) and v(x,y), holomorphicity on an open set in ℂ is equivalent to satisfying, at every point, the system ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x. That equivalence matters because it provides a concrete, calculable test for complex differentiability using ordinary multivariable calculus.

The first example is the identity function f(z)=z. Writing z=x+iy gives f(x+iy)=x+iy, so the real part is u(x,y)=x and the imaginary part is v(x,y)=y. Computing derivatives is immediate: ∂u/∂x=1 and ∂v/∂y=1, so the first Cauchy–Riemann equation holds everywhere. For the second equation, ∂u/∂y=0 while ∂v/∂x=0, making −∂v/∂x also 0. Both equations are satisfied at every point, so f(z)=z is holomorphic.

The next example flips the conclusion by using complex conjugation, f(z)=z. With z=x+iy, conjugation yields f(x+iy)=x−iy, so u(x,y)=x and v(x,y)=−y. The first Cauchy–Riemann equation still matches: ∂u/∂x=1 and ∂v/∂y=−1? Actually ∂v/∂y=−1, so the equality ∂u/∂x = ∂v/∂y fails. The mismatch persists across the domain, leading to the expected result: complex conjugation is not holomorphic.

A more involved polynomial example then demonstrates how the same test works even when u and v are not obvious at first glance. Consider f(z)=z^2+i z. Substituting z=x+iy and expanding gives a real part u(x,y) and an imaginary part v(x,y) after collecting terms: u(x,y)=x^2−y^2−y and v(x,y)=2xy+x (as obtained by grouping the non-i terms and the i-multiplied terms). Checking Cauchy–Riemann requires four derivatives: ∂u/∂x=2x and ∂v/∂y=2x, so the first equation holds. For the second, ∂u/∂y=−2y−1 and ∂v/∂x=2y+1, so −∂v/∂x=−(2y+1)=−2y−1, matching ∂u/∂y. With both equations satisfied everywhere, f(z)=z^2+i z is holomorphic.

Overall, the examples show a practical workflow: rewrite f in terms of x and y, identify u and v, compute the relevant partial derivatives, and verify the Cauchy–Riemann equalities. The payoff is immediate classification—holomorphic for the identity and the polynomial, non-holomorphic for conjugation—without needing to compute complex derivatives directly.

Cornell Notes

Holomorphicity of a complex function f=u+iv on an open set in ℂ is equivalent to satisfying the Cauchy–Riemann equations at every point: ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x. The identity f(z)=z becomes u=x and v=y, and both equations hold (1=1 and 0=0), so it is holomorphic. Complex conjugation f(z)=z becomes u=x and v=−y, causing ∂u/∂x=1 to disagree with ∂v/∂y=−1, so it is not holomorphic. For f(z)=z^2+i z, expanding in x and y yields u and v; computing the four partial derivatives shows both Cauchy–Riemann equations match, confirming holomorphicity.

How does one apply the Cauchy–Riemann equations to a given complex function f(z)?

Write z=x+iy and rewrite f(x+iy) in the form u(x,y)+i v(x,y). Then compute ∂u/∂x and ∂v/∂y and check whether they are equal. Next compute ∂u/∂y and ∂v/∂x and check whether ∂u/∂y = −∂v/∂x. If both equalities hold at every point in the open domain, f is holomorphic.

Why is f(z)=z holomorphic under the Cauchy–Riemann test?

For f(z)=z, substituting z=x+iy gives f=x+iy, so u(x,y)=x and v(x,y)=y. Then ∂u/∂x=1 and ∂v/∂y=1, satisfying the first equation. Also ∂u/∂y=0 and ∂v/∂x=0, so ∂u/∂y = −∂v/∂x becomes 0=−0, satisfying the second equation everywhere.

What exactly fails for complex conjugation f(z)=z?

With z=x+iy, conjugation gives f=x−iy, so u(x,y)=x and v(x,y)=−y. The first Cauchy–Riemann equation requires ∂u/∂x = ∂v/∂y. Here ∂u/∂x=1, but ∂v/∂y=−1, so the equality fails at every point. That single sign flip in v is enough to rule out holomorphicity.

How do you handle a function where u and v aren’t obvious, like f(z)=z^2+i z?

Expand after substituting z=x+iy. Collect all terms without i into u(x,y) and all terms multiplied by i into v(x,y). For f(z)=z^2+i z, expansion and grouping produce u(x,y)=x^2−y^2−y and v(x,y)=2xy+x. Then compute ∂u/∂x=2x and ∂v/∂y=2x (first equation), and compute ∂u/∂y=−2y−1 and ∂v/∂x=2y+1 so that −∂v/∂x=−2y−1 (second equation). Both match, confirming holomorphicity.

What does it mean when one Cauchy–Riemann equation holds but the other doesn’t?

Holomorphicity requires both equations simultaneously. If ∂u/∂x = ∂v/∂y holds but ∂u/∂y = −∂v/∂x fails (or vice versa), the function cannot be complex differentiable in the holomorphic sense on that domain. The conjugation example illustrates this: the sign change in v makes the first equality fail everywhere.

Review Questions

Given f(z)=u(x,y)+i v(x,y), what two derivative equalities must hold for holomorphicity?
For f(z)=z, compute u and v and show which Cauchy–Riemann equation fails.
For f(z)=z^2+i z, identify u(x,y) and v(x,y), then verify both Cauchy–Riemann equations by computing the needed partial derivatives.

Key Points

1
Write any complex function as f(z)=u(x,y)+i v(x,y) by substituting z=x+iy and separating real and imaginary parts.
2
Holomorphicity on an open set is equivalent to satisfying ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x at every point.
3
For f(z)=z, u=x and v=y, and both Cauchy–Riemann equations hold everywhere.
4
For complex conjugation f(z)=z, u=x and v=−y, causing ∂u/∂x=1 to disagree with ∂v/∂y=−1, so the function is not holomorphic.
5
When u and v are not immediate, expand f(x+iy) and collect terms to isolate u (non-i terms) and v (i-multiplied terms).
6
For f(z)=z^2+i z, expanding and grouping yields u(x,y)=x^2−y^2−y and v(x,y)=2xy+x, and the Cauchy–Riemann equations both check out.
7
The Cauchy–Riemann test provides a direct multivariable-calculus method to classify functions as holomorphic or not without computing complex derivatives directly.

Highlights

Complex holomorphicity becomes a pair of partial differential equations: ∂u/∂x = ∂v/∂y and ∂u/∂y = −∂v/∂x.

Identity f(z)=z passes the test because u=x and v=y make all required derivative equalities match.

Complex conjugation fails immediately due to the sign change in the imaginary part v=−y, flipping ∂v/∂y.

For f(z)=z^2+i z, careful expansion into u and v lets the Cauchy–Riemann equations verify holomorphicity through matching derivatives.

Topics

Cauchy-Riemann Equations
Holomorphic Functions
Complex Differentiability
Examples
Real and Imaginary Parts