2.7: Cauchy-Riemann all the way down

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We’ve defined an analytic function as one having a complex derivative. The following theorem shows that if \(f\) is analytic then so is \(f'\). Thus, there are derivatives all the way down!

Theorem \(\PageIndex{1}\)

Assume the second order partials of \(u\) and \(v\) exist and are continuous. If \(f(z) = u + iv\) is analytic, then so is \(f'(z)\).

Proof

To show this we have to prove that \(f'(z)\) satisfies the Cauchy-Riemann equations. If \(f = u + iv\) we know

\(u_x = v_y\), \(u_y = -v_x\), \(f' = u_x + iv_x\).

Let's write

\[f' = U + iV, \nonumber \]

so, by Cauchy-Riemann,

\[U = u_x = u_y,\ \ V= v_x = -u_y. \nonumber \]

We want to show that \(U_x = V_y\) and \(U_x = -V_x\). We do them one at a time.

To prove \(U_x = V_y\), we use Equation 2.8.2 to see that

\[U_x = v_{yx} \text{ and } V_y = v_{xy}. \nonumber \]

Since \(v_{xy} = v_{yx}\), we have \(U_x = V_y\).

Similarly, to show \(U_y = -V_x\), we compute

\[U_y = u_{xy} \text{ and } V_x = -u_{yx}. \nonumber \]

So, \(U_y = -V_x\). QED.

Technical point. We’ve assumed as many partials as we need. So far we can’t guarantee that all the partials exist. Soon we will have a theorem which says that an analytic function has derivatives of all order. We’ll just assume that for now. In any case, in most examples this will be obvious.