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Post by mondo on Aug 12, 2022 5:33:28 GMT
On page 515, in the 3rd paragraph we have $\Phi(z) = \frac{S}{4}|z|^2$ and author says it generates $V = \frac{S}{2}z$. I believe the calculations he did were $\frac{d}{dx}(\frac{S}{4}(x^2 + y^2)) + \frac{d}{dy}(\frac{S}{4}(x^2 + y^2))$ hence we have $\frac{S}{4}(2x) + \frac{S}{4}(2y) = \frac{S}{2}(x+y))$ So it appears author replaced $(x+y)$ with $z$ but is it "legal" here? After all $(x+y) \neq $(x+iy) = z$. Another doubt comes from the fact that we were differentiating a real number $|z|^2$, a length of our complex vector $z$ squared and we got a complex number back? That doesn't sound right.
Thank you.
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Post by Admin on Aug 12, 2022 5:56:09 GMT
On page 515, in the 3rd paragraph we have $\Phi(z) = \frac{S}{4}|z|^2$ and author says it generates $V = \frac{S}{2}z$. I believe the calculations he did were $\frac{d}{dx}(\frac{S}{4}(x^2 + y^2)) + \frac{d}{dy}(\frac{S}{4}(x^2 + y^2))$ hence we have $\frac{S}{4}(2x) + \frac{S}{4}(2y) = \frac{S}{2}(x+y))$ So it appears author replaced $(x+y)$ with $z$ but is it "legal" here? After all $(x+y) \neq $(x+iy) = z$. Another doubt comes from the fact that we were differentiating a real number $|z|^2$, a length of our complex vector $z$ squared and we got a complex number back? That doesn't sound right. Thank you. Mondo You have got the definition of $\nabla\Phi$ wrong as you did the other week. This changes a scalar into a vector, Remember? I pointed this out to you recently. This book does not contain very many errors and most of them are documented in the forum. If you find you don't understand something the chances are that you need to think a bit harder. I always assume that I am wrong if I don't understand something in the book. Vasco
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Post by mondo on Aug 12, 2022 6:11:10 GMT
Ahh so,
Should rather be $\frac{S}{4}(2x) + i\frac{S}{4}(2y) = \frac{S}{2}(x+iy) = \frac{S}{2}(z)$
Now it makes sense, thank you Vasco.
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