Question

Show that the function f(z) = x^2 + y^2 + i 2xy has a derivative only at points that lie on

the x-axis.

Answer 1

$u = {\mathop{\rm Re}\nolimits} f(z) = {x^2} + {y^2},\,\,v = {\mathop{\rm Im}\nolimits} f(z) = 2xy$

Then

$\frac{{\partial u}}{{\partial x}} = 2x;\,\frac{{\partial u}}{{\partial y}} = 2y,\,\frac{{\partial v}}{{\partial x}} = 2y,\,\frac{{\partial v}}{{\partial y}} = 2x$

We write down the Cauchy-Riemann conditions:

$\left\{ \begin{array}{l} \frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}\\ \frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}} \end{array} \right.$

$\left\{ \begin{array}{l} 2x = 2x\\ 2y = - 2y \end{array} \right.$

From 2-nd equation we have $y=0$ .

But then the point for which the Cauchy-Riemann conditions are satisfied must lie on the x-axis, which means that the function is differentiable only on this axis. Q.E.D.