A student wants to find the harmonic conjugate of u = (1/2)ln(x² + y²) on the punctured plane ℝ² \ {0}. They integrate the Cauchy-Riemann equations and obtain v = arctan(y/x), concluding this is the harmonic conjugate. What is wrong with this reasoning?
AThe Cauchy-Riemann equations only apply to analytic functions, not to arbitrary harmonic functions
BThe integration is incorrect — the harmonic conjugate of (1/2)ln(x² + y²) is |z|, not arctan(y/x)
CThe punctured plane is not simply connected, so the integral is path-dependent and arctan(y/x) is multivalued — its value changes by 2π when looping around the origin
DThe function u = (1/2)ln(x² + y²) is not harmonic on the punctured plane
The Cauchy-Riemann integration is formally correct — it yields arctan(y/x) as the 'natural' candidate. But the punctured plane ℝ² \ {0} is not simply connected: it has a hole at the origin. When you traverse a closed loop around the origin, arctan(y/x) increases by 2π rather than returning to its starting value. This multivaluedness disqualifies arctan(y/x) as a harmonic conjugate, which must be a single-valued function. On a simply connected domain that excludes the origin (like the right half-plane), the conjugate does exist — but no simply connected domain can cover the full punctured plane.
Question 2 Multiple Choice
To find the harmonic conjugate of u = x³ − 3xy², what is the correct procedure?
ASet v = −(y³ − 3x²y) and verify the Cauchy-Riemann equations hold
BIntegrate ∂v/∂y = ∂u/∂x = 3x² − 3y² with respect to y, obtaining v = 3x²y − y³ + g(x), then use ∂v/∂x = −∂u/∂y to determine g(x)
CTake the imaginary part of eᶻ where the real part equals u
DDifferentiate u twice and solve Laplace's equation directly for v
The standard construction: since ∂v/∂y = ∂u/∂x = 3x² − 3y², integrate with respect to y to get v = 3x²y − y³ + g(x), where g(x) is an unknown function of x alone. Then apply the second Cauchy-Riemann equation: ∂v/∂x = 6xy + g'(x) must equal −∂u/∂y = −(−6xy) = 6xy. So g'(x) = 0, meaning g is a constant. The harmonic conjugate is v = 3x²y − y³ + C, and the corresponding holomorphic function is f = (x³ − 3xy²) + i(3x²y − y³) = z³.
Question 3 True / False
Nearly every harmonic function defined on any open connected domain in ℝ² has a harmonic conjugate.
TTrue
FFalse
Answer: False
This is only true on simply connected domains — domains with no holes. On domains that are not simply connected (like the punctured plane, or an annulus), a harmonic function may not have a single-valued harmonic conjugate because the line integral used to construct it can be path-dependent. The canonical counterexample is u = (1/2)ln(x² + y²) on ℝ² \ {0}: it is harmonic there, but its 'natural' conjugate arctan(y/x) is multivalued. On any simply connected domain, the Cauchy-Riemann equations can always be integrated to produce a single-valued harmonic conjugate.
Question 4 True / False
If v is the harmonic conjugate of u on a simply connected domain, then u is also the harmonic conjugate of v, and both u and v individually satisfy Laplace's equation.
TTrue
FFalse
Answer: True
Both claims follow from the structure of holomorphic functions. If f = u + iv is holomorphic, then if = −v + iu is also holomorphic (multiplication by i rotates the function). This shows that u is the real part of a holomorphic function whose imaginary part is −v, confirming the symmetric relationship: u and v are each other's harmonic conjugates (up to sign). Additionally, both u and v are the real and imaginary parts of a holomorphic function, and all such parts satisfy Laplace's equation — this is a direct consequence of the Cauchy-Riemann equations.
Question 5 Short Answer
Why does the harmonic conjugate of u = (1/2)ln(x² + y²) fail to exist on the punctured plane ℝ² \ {0}, even though u is harmonic there?
Think about your answer, then reveal below.
Model answer: The punctured plane is not simply connected — it has a hole at the origin. The 'natural' conjugate arctan(y/x) is multivalued: traveling along a closed loop around the origin returns to the same point but shifts arctan's value by 2π. A harmonic conjugate must be single-valued, so no harmonic conjugate exists on the full punctured plane.
On a simply connected domain, any closed path can be continuously contracted to a point without crossing the boundary, which guarantees that line integrals are path-independent and the Cauchy-Riemann construction produces a single-valued function. On the punctured plane, paths that wind around the origin cannot be contracted — they are topologically distinct, and the integral accumulates an extra 2π per winding. This is not a failure of the Cauchy-Riemann equations but a topological obstruction. The remedy is to restrict to a simply connected subset: on the right half-plane (x > 0), for example, the branch of arctan that maps into (−π/2, π/2) is a valid single-valued harmonic conjugate.