 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2023/3/a/solution

At every

z = x + i y \in U

u_{x} = v_{y}, u_{y} = - v_{x} .

To derive them from complex differentiability, approach

z

along real increments

h

f^{'} (z) = lim_{h \to 0} \frac{f ( z + h ) - f ( z )}{h} = u_{x} + i v_{x} .

Approaching along imaginary increments

ih

instead gives

f^{'} (z) = lim_{h \to 0} \frac{f ( z + ih ) - f ( z )}{ih} = v_{y} - i u_{y} .

These limits exist because

f

is holomorphic. Equating their real and imaginary parts proves the equations, together with

f^{'} = u_{x} + i v_{x} = v_{y} - i u_{y}

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