# Cauchy-Riemann Equations problem (f(z) = ze^-z)

## Main Question or Discussion Point

This is not a homework problem - it was in a review book that I am studying before finals. The problem stated: Verify the Cauchy-Riemann equations for the following functions. Deduce that they are analytic. The function I am having trouble with is $$f(z) = z e^{-z}$$

Now, before I show my work, I just want to say that it seems quite obvious that this function is analytic, without the use of the Cauchy-Riemann functions.

$$g(z) = z$$, the identity map, is entire.

$$h(z) = e^{z}$$ is also entire, and never equal to zero.

Therefore, since $$f(x) = ze^{-z} = \frac{g(z)}{h(z)}$$ is the quotient of two entire functions (and the denominator is never 0) it must also be entire.

I also checked that the Cauchy-Riemann equations work for both functions $$g$$ and $$\frac{1}{h}$$ individually, but when I multiply them together, it doesn't work out. Where's my mistake?

$$z = x + iy$$

$$ze^{-z} = (x + iy)(e^{-x}(cos y - i sin y)) = x e^{-x} cos y + y sin y + i (y e^{-x} cos y - x sin y)$$

So, let the real part of this function be $$U(x,y)$$ and let the imaginary part be $$V(x,y)$$.

The Cauchy-Riemann equations say that $$U_x = V_y$$ and $$U_y = - V_x$$. But I get:

$$U_x = -x e^{-x} cos y + e^{-x} cos y$$

$$V_y = -y e^{-x} sin y + e^{-x} cosy - x cos y$$

Clearly, these aren't the same...

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The exponent $e^{-x}$ should be a common factor for all of your terms.. but in the line where you multiply everything out, you seem to have dropped it from some terms..

$$ze^{-z} = (x + iy)(e^{-x}(cos y - i sin y)) \neq x e^{-x} cos y + y sin y + i (y e^{-x} cos y - x sin y)$$

Oh, I see... duh.

Thanks!

mathman