# Show that `f(z) = z |z|`

is not analytic anywhere in the complex plane

To show that `f(z) = z |z|`

is not analytic anywhere in the complex plane, we can use the Cauchy-Riemann equations. A function `f(z) = u(x, y) + iv(x, y)`

is analytic if and only if the Cauchy-Riemann equations are satisfied and the partial derivatives of `u`

and `v`

are continuous.

First, express `f(z) = z |z|`

in terms of `z = x + iy`

:

|z| = sqrt(x^{2}+ y^{2})

f(z) = (x + iy) sqrt(x^{2}+ y^{2})

Let `u(x, y)`

and `v(x, y)`

be the real and imaginary parts of `f(z)`

:

u(x, y) = x sqrt(x^{2}+ y^{2})

v(x, y) = y sqrt(x^{2}+ y^{2})

The Cauchy-Riemann equations are:

∂u/∂x = ∂v/∂y

∂u/∂y = -∂v/∂x

Now, compute the partial derivatives of `u`

and `v`

:

∂u/∂x = ∂/∂x (x sqrt(x^{2}+ y^{2}))

Using the product rule:

∂u/∂x = sqrt(x^{2}+ y^{2}) + x ∙ ∂/∂x (sqrt(x^{2}+ y^{2}))

∂/∂x (sqrt(x^{2}+ y^{2})) = x/sqrt(x^{2}+ y^{2})

Thus,

∂u/∂x = sqrt(x^{2}+ y^{2}) + x ∙ x/sqrt(x^{2}+ y^{2})

∂u/∂x = sqrt(x^{2}+ y^{2}) + x^{2}/sqrt(x^{2}+ y^{2})

∂u/∂x = (x^{2}+ y^{2}+ x^{2})/sqrt(x^{2}+ y^{2})

∂u/∂x = (2x^{2}+ y^{2})/sqrt(x^{2}+ y^{2})

Next, compute `∂v/∂y`

:

∂v/∂y = ∂/∂y (y sqrt(x^{2}+ y^{2}))

Using the product rule:

∂v/∂y = sqrt(x^{2}+ y^{2}) + y ∙ ∂/∂y (sqrt(x^{2}+ y^{2}))

∂/∂y (sqrt(x^{2}+ y^{2})) = y/sqrt(x^{2}+ y^{2})

Thus,

∂v/∂y = sqrt(x^{2}+ y^{2}) + y ∙ y/sqrt(x^{2}+ y^{2})

∂v/∂y = sqrt(x^{2}+ y^{2}) + y^{2}/sqrt(x^{2}+ y^{2})

∂v/∂y = (x^{2}+ y^{2}+ y^{2})/sqrt(x^{2}+ y^{2})

∂v/∂y = (x^{2}+ 2y^{2})/sqrt(x^{2}+ y^{2})

For the Cauchy-Riemann equations to hold, we must have:

∂u/∂x = ∂v/∂y

However,

(2x^{2}+ y^{2})/sqrt(x^{2}+ y^{2}) ≠ (x^{2}+ 2y^{2})/sqrt(x^{2}+ y^{2})

Since the Cauchy-Riemann equations are not satisfied, `f(z) = z |z|`

is not analytic anywhere in the complex plane.