# A Harmonic Function is Zero on an open portion of the boundary, help!

A harmonic function in a region is zero on an open portion of the boundary, and its normal derivative is also zero on the same part, and it is continuously differentiable on the boundary. I have to show that the function is zero everywhere, but I have no idea how. I have tried this for hours and hours, and haven't come up with anything useful. The best answer I've had so far involves the Cauchy Kovelevskaya theorem, but that was shown to be flawed. Can anyone help? This is very difficult..

I'm pretty sure this implies the gradient is zero on the boundary, is there anything I can possible do with that?

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