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

[BSMSMSTMSPHD]

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...

Thanks for your help!

 

[jpr0]

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)

 

[BSMSMSTMSPHD]

Oh, I see... duh.

Thanks!

 

[mathman]

In your equation for ze^-z, your expressions for U and V are incorrect. e^-x multiplies all terms.