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I have found that:

$U_x = -\exp(y)\sin(x) $

$U_y = \exp(y)\cos(x) $

$V_x = \exp(y)\cos(x) $

$V_y = \exp(y)\sin(x) $

I need to show that $U_x=V_y$ and $U_y=-V_x$, however these aren't satisfied? does that mean $f$ is not differentiable?

Michael Hardy
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nep
  • 21

1 Answers1

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Correct. If the same expression that appears on the right side were $f(y+ix)$ rather than $f(x+iy)$ then it would be differentiable.

Michael Hardy
  • 1
  • hypothetically in this case if we could set $x$ or $y$ to be such that both equations were satisfied then am I right to assume it will be differentiable at those points – nep Apr 15 '15 at 21:44