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## Main Question or Discussion Point

**how to show a function is analytic??**

I know that to show a function is analytic I need to prove its differentiable, but for a fuction like log(z-i), how could i show it is analytic???

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- #1

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I know that to show a function is analytic I need to prove its differentiable, but for a fuction like log(z-i), how could i show it is analytic???

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HallsofIvy

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Use the "Cauchy-Riemann equations which should be mentioned early in any book on "functions of a complex variable". A function f(x+ iy)= u(x,y)+ iv(x,y) is analytic at [itex]z_0= x_0+ iy_0[/itex] if and only if the partial derivatives, [itex]\partial u/\partial x[/itex], [itex]\partial u/\partial y[/itex], [itex]\partial v/\partial x[/itex], and [itex]\partial v/\partial y[/itex] are continous at the point and

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

and

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

- #3

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yes, i know wat u mean, but i dont know how to seperate log(z-i) into u + iv form

- #4

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Why not just differentiate it and then show the derivative exists in a region surrounding a point then it is analytic in that region so:I know that to show a function is analytic I need to prove its differentiable, but for a fuction like log(z-i), how could i show it is analytic???

[tex]\frac{d}{dz} \log(z-i)=\frac{1}{z-i}[/tex]

and that derivative exists everywhere except at z=i.

- #5

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Or you could integrate the function over a closed line and show the integral is zero.

- #6

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use log(z) = log(|z|) + i (arg(z))

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