- #26

Ben Niehoff

Science Advisor

Gold Member

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Returning to the Cauchy-Riemann equations, we can consider the complex exponential to be a function

[tex]f(x,y) = u(x,y) + i v(x,y)[/tex]

such that f, u, v, x, and y are real, and f satisfies

[tex]\frac{df}{dz} = f[/tex]

where [itex]z = x+iy[/itex].

I don't have time to work this out right now, but this will lead to two coupled partial differential equations for u and v. Combined with the equations

[tex]\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}[/tex]

[tex]\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}[/tex]

we should be able to show that

[tex]u(x,y) = C \exp x \cos y[/tex]

[tex]v(x,y) = C \exp x \sin y[/itex]

for arbitrary real constant C. Then we simply choose the solution that is consistent with [itex]\exp x[/itex] on the real line.

Hmm...no, I think C should turn out to be complex...but there is still a unique solution that is consistent with the real exponential.

[tex]f(x,y) = u(x,y) + i v(x,y)[/tex]

such that f, u, v, x, and y are real, and f satisfies

[tex]\frac{df}{dz} = f[/tex]

where [itex]z = x+iy[/itex].

I don't have time to work this out right now, but this will lead to two coupled partial differential equations for u and v. Combined with the equations

[tex]\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}[/tex]

[tex]\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}[/tex]

we should be able to show that

[tex]u(x,y) = C \exp x \cos y[/tex]

[tex]v(x,y) = C \exp x \sin y[/itex]

for arbitrary real constant C. Then we simply choose the solution that is consistent with [itex]\exp x[/itex] on the real line.

Hmm...no, I think C should turn out to be complex...but there is still a unique solution that is consistent with the real exponential.

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