Author Topic: The Partial Derivative of f = u + iv on textbook  (Read 681 times)

Yuhan Shao

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The Partial Derivative of f = u + iv on textbook
« on: October 16, 2018, 10:01:01 PM »
The attachments are two different partial derivatives of f = u + iv respective to x and y
Is there a typo or I am missing something?
The one in the 2.1 seems correct.

Thanks!

Min Gyu Woo

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Re: The Partial Derivative of f = u + iv on textbook
« Reply #1 on: October 16, 2018, 10:29:54 PM »
It seems like the equation in 2.1 is

$$\frac{df(z)}{dz}$$

while the equation in 1.6 is

$$\frac{\partial f(p,q)}{\partial x} \text{ and } \frac{\partial f(p,q)}{\partial y}$$

Basically, the 2.1 equation talks about the derivative while 1.6 equations talk about the two separate partial derivatives.

oighea

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Re: The Partial Derivative of f = u + iv on textbook
« Reply #2 on: October 16, 2018, 10:37:28 PM »
There is no typo. Also, we can view $u$ as $\mathfrak{Re}f$ $v$ as $\mathfrak{Im}f$. The equation in 2.1 is the total derivative, whereas the equation in 1.6 are the two partial derivatives, one in respect to the real values $\frac{\delta u}{\delta x} + i\frac{\delta v}{\delta x}$ and the other in respect to the imaginary values $\frac{\delta u}{i\delta y} + \frac{+\delta v}{\delta y}$ (note that in the limit definition, you have to watch out for the $i$ on the denominator $k$).

The total derivative can be expressed as $\frac{df}{dz} = \frac{\delta u}{\delta x}\frac{dx}{dz} + i\frac{\delta v}{\delta x}\frac{dx}{dz} - i\frac{\delta u}{\delta y}\frac{dy}{dz} + \frac{\delta v}{\delta y}\frac{dy}{dz}$