Sep 1, 2011 #1 demidemi Messages 3 Reaction score 0 Homework Statement If z=x + iy, what is d/dz in polar coordinates? The Attempt at a Solution I know that expanded, d/dz = 1/2 (d/dx) - i (d/dy) Where to go from there?
Homework Statement If z=x + iy, what is d/dz in polar coordinates? The Attempt at a Solution I know that expanded, d/dz = 1/2 (d/dx) - i (d/dy) Where to go from there?
Sep 2, 2011 #2 HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 984 Use the chain rule: You have \frac{df}{dz}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}i = \left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\right)+ \left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial y}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y}\right)i with, of course, r= \sqrt{x^2+ y^2} and \theta= arctan(y/x).
Use the chain rule: You have \frac{df}{dz}= \frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}i = \left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial f}{\partial \theta}\right)+ \left(\frac{\partial f}{\partial r}\frac{\partial r}{\partial y}+ \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y}\right)i with, of course, r= \sqrt{x^2+ y^2} and \theta= arctan(y/x).
Sep 4, 2011 #3 demidemi Messages 3 Reaction score 0 Why is "f" necessary there? Also, should there be a "partial theta partial x" in the first parentheses? Last edited: Sep 4, 2011
Why is "f" necessary there? Also, should there be a "partial theta partial x" in the first parentheses?