Let $f(z) = u(x,y) + iv(x,y)$ be a complex function that is differentiable at the point $z_0 =x_0 + iy_0$. Prove that $f'(z_0)= \frac{\partial u}{\partial x} (x_0,y_0) + i \frac{\partial v}{\partial x} (x_0,y_0)$. I have been at this for a while now but ii do not know how to approach this problem. Is that statement even true. I was thinking about the Cauchy rieman equations to see if there is a relation here i could not find nothing. Then i use the definition of derivative but this not right because the function is differentiable at a point so it has to be Cauchy. Anyone help please.
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2 Answers
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1 If our function $f$ is differentiable at $z_0$, then $$ \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z - z_0} $$ exists. For this limit to exist, it must exist regardless of how $z$ approaches $z_0$. In particular, we can consider how $z = x + iy$ approaches $z_0$ as $y = y_0$ is fixed, and $x$ varies. Since we assume the above limit exists, we have $$ f'(z_0) = \lim_{x \to x_0} \frac{u(x, y_0) + iv(x, y_0) - (u(x_0, y_0) + iv(x_0, y_0)}{(x + iy_0) - (x_0 + iy_0)}. $$ But you can break this fraction into the sum of two fractions, one involving the function $u$ and one involving the function $v$. From here you will get the desired result.
- $\begingroup$ oh i see thank man i appreciate it $\endgroup$user146269– user1462692015-04-23 21:30:48 +00:00Commented Apr 23, 2015 at 21:30
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3 By definition, $f^{\prime}(z_0) = \mathrm{lim}_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0} $
If the function is differentiable, then the limit exists. If the limit exists, then it is independent of how $z \to z_0$. In particular, we could fix $y_0$ and approach along $x$.
- $\begingroup$ i know that but what i am really asking is how do you manipulate $f'(z)$ in order to see the cauchy rieman $\endgroup$user146269– user1462692015-04-23 21:30:09 +00:00Commented Apr 23, 2015 at 21:30
- $\begingroup$ If you want to get Cauchy-Riemann out of $f^{\prime}$, first evaluate by approaching along $x$, then do the same thing but approaching along $y$. You'll get two expressions that have to be equal, so their real and imaginary parts have to be equal as well. $\endgroup$Tom– Tom2015-04-23 21:32:46 +00:00Commented Apr 23, 2015 at 21:32
- $\begingroup$ cool thanks guys i appreciate it $\endgroup$user146269– user1462692015-04-23 21:33:34 +00:00Commented Apr 23, 2015 at 21:33