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Post by Admin on Jul 13, 2022 8:12:39 GMT
Mondo
Glad you made progress and happy to help.
Not sure what you mean by complex vectors?? in your post above.
Complex numbers are vectors. If you sum complex numbers it's the same as summing vectors.
Is this just a question of how you understand the words and how I understand the words, or are you still a bit confused?
Vasco
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Post by Admin on Jul 13, 2022 11:02:11 GMT
Mondo
Using the mid-point Riemann sum for the real integral $W$ when $\overline{H}=z$ and $ds=dy=0.5$ gives:
$\displaystyle W=\int_K(\text{Tangential component of }\overline{H})ds= \int_{y=-1}^{y=+1}ydy=\delta y(y_1+y_2+y_3+y_4)=(1/2)(-3/4-1/4+1/4+3/4)=0$ where
$z_1=x_1+iy_1=1-(3/4)i$,
$z_2=x_2+iy_2=1-(1/4)i$,
$z_3=x_3+iy_3=1+(1/4)i$,
$z_4=x_4+iy_4=1+(3/4)i$
If we use the Riemann sum for the complex integral on the LHS of (11) then since $H=\overline{z}$, we would write
$\displaystyle\int_KHdz=i(1/2)[(x_1-iy_1)+(x_2-iy_2)+(x_3-iy_3)+(x_4-iy_4)]$
$=(1/2)[(ix_1+y_1)+(ix_2+y_2)+(ix_3+y_3)+(ix_4+y_4)]$
$=(1/2)(y_1+y_2+y_3+y_4)+(1/2)i(x_1+x_2+x_3+x_4)$
$=0+2i$
Vasco
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Post by mondo on Jul 14, 2022 6:08:03 GMT
Not sure what you mean by complex vectors?? in your post above. Complex numbers are vectors. If you sum complex numbers it's the same as summing vectors. Not quite, they can only be understood as 2D vectors but vectors in general can be $n$ dimensional which can not be said about complex numbers, right? Is this just a question of how you understand the words and how I understand the words, or are you still a bit confused? A bit confused with one more thing - I feel like we have a collision in naming convention here. $(1-i)$ is a vector you say, I agree but then how would we call a single vector in a vector field defined by $\bar H(z) = |z|^2z $? This is also a vector that originates at $z$ and terminates at $|z|^2z$. That said I have no problem visualizing work integral for $\bar H(z) = |z|^2z$ but I have a problem with $\bar H(z) = z$ because our vector here is just a point! So, first of all I think our mid point Riemann's sum just doesn't belong here - this type of integral won't work for vector fields. Second, since for $\bar H(z) = z $ we have vectors of $0$ length the integral $\int_{-1}^{1} Im(z)dy$ would give us a correct result but is a bit misleading as to how this vector fields actually look like. I mean, looking at the sum itself it seems like we add vectors of imaginary length while in actuality we add 0 length vectors. ( Or I still miss sth ) So that said: Mondo Using the mid-point Riemann sum for the real integral $W$ when $\overline{H}=z$ and $ds=dy=0.5$ gives: $\displaystyle W=\int_K(\text{Tangential component of }\overline{H})ds= \int_{y=-1}^{y=+1}ydy=\delta y(y_1+y_2+y_3+y_4)=(1/2)(-3/4-1/4+1/4+3/4)=0$ where Gives correct answer but I would say using incorrect method since we can't treat vector field as midpoints. Do you agree Vasco? |
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Post by Admin on Jul 14, 2022 8:40:01 GMT
Not sure what you mean by complex vectors?? in your post above. Complex numbers are vectors. If you sum complex numbers it's the same as summing vectors. Not quite, they can only be understood as 2D vectors but vectors in general can be $n$ dimensional which can not be said about complex numbers, right? Mondo
What I wrote is correct. Complex numbers are vectors. I agree that it is not correct to say vectors are complex numbers, but that's not what I said.
Complex numbers are vectors. This is logically correct.
Men are humans - correct, humans are men - incorrect. They may be women.Is this just a question of how you understand the words and how I understand the words, or are you still a bit confused? A bit confused with one more thing - I feel like we have a collision in naming convention here. $(1-i)$ is a vector you say, I agree but then how would we call a single vector in a vector field defined by $\bar H(z) = |z|^2z $? This is also a vector that originates at $z$ and terminates at $|z|^2z$. That said I have no problem visualizing work integral for $\bar H(z) = |z|^2z$ but I have a problem with $\bar H(z) = z$ because our vector here is just a point! So, first of all I think our mid point Riemann's sum just doesn't belong here - this type of integral won't work for vector fields. Second, since for $\bar H(z) = z $ we have vectors of $0$ length the integral $\int_{-1}^{1} Im(z)dy$ would give us a correct result but is a bit misleading as to how this vector fields actually look like. I mean, looking at the sum itself it seems like we add vectors of imaginary length while in actuality we add 0 length vectors. ( Or I still miss sth ) Yes you are missing something. What you are missing is chapter 10 section 1 on pages 450-1 Make sure you do the exercises suggested using equation explorer.
When the vector field is $z$ it is a bit of a special case: you have the vector $z$ at the point $z$. The vector $z$ is not a vector of zero length! It is a vector of length $|z|$
When you say "$z$ is just a point" you need to realise that any complex quantity can be thought of as a point in the complex plane or as a vector in the complex plane. So that said: Mondo Using the mid-point Riemann sum for the real integral $W$ when $\overline{H}=z$ and $ds=dy=0.5$ gives: $\displaystyle W=\int_K(\text{Tangential component of }\overline{H})ds= \int_{y=-1}^{y=+1}ydy=\delta y(y_1+y_2+y_3+y_4)=(1/2)(-3/4-1/4+1/4+3/4)=0$ where Gives correct answer but I would say using incorrect method since we can't treat vector field as midpoints. Do you agree Vasco? No I don't agree. Please please please read the beginning of chapter 10. When $H(z)$ is a vector field defined everywhere in the complex plane every point of the complex plane has the appropriate vector $H$ emanating from it.
I am not treating the vector field as mid points. We have mid points where the vector field (chosen by you) just happens to be $z$.
Mid points are points in the complex plane. The vector field is defined as having a value at each point of the plane.Vasco |
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Post by mondo on Jul 16, 2022 20:30:20 GMT
Vasco, thank you for a response. Before I answer your comments from previous post can we clarify this - why for a vector field $H(z) = z^2$ (I plotted it here myminiurl.net/befXT together with helper lines at $y=3$ and $x=-9$) when I calculate $H(3i)$ I get $-9$. Hence I would expect a vector from $3i$ going down to $-9$ on a negative real axis. Instead what I see is a short horizontal vector pointing in a negative real axis. I use the question $H(z) = z^2 = (a+ib)^2 = (a^2-b^2) + 2aib$. |
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Post by Admin on Jul 16, 2022 21:01:25 GMT
Vasco, thank you for a response. Before I answer your comments from previous post can we clarify this - why for a vector field $H(z) = z^2$ (I plotted it here myminiurl.net/befXT together with helper lines at $y=3$ and $x=-9$) when I calculate $H(3i)$ I get $-9$. Hence I would expect a vector from $3i$ going down to $-9$ on a negative real axis. Instead what I see is a short horizontal vector pointing in a negative real axis. I use the question $H(z) = z^2 = (a+ib)^2 = (a^2-b^2) + 2aib$. Mondo It's because in your plot you are using a scale factor equal to $0.1$ and so $-9$ is plotted as $-0.9$ Vasco |
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Post by mondo on Jul 16, 2022 23:31:55 GMT
![]() Hmm but this vector is horizontal while it should head down to -9. Also, I ploted it using wolfram:  aand here the scale is from -10 to 10 and the vector at $3i$ still looks very tiny. |
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Post by Admin on Jul 16, 2022 23:54:43 GMT
Mondo
No it is horizontal.
-9 is a vector with no vertical component ie horizontal.
You don't understand vectors or vector fields.
Study them some more. Have you seen the figure at the beginning of chapter 10 of the vector field $z^2$?
Not the scale the scale factor
Vasco
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Post by Admin on Jul 17, 2022 0:10:55 GMT
Mondo
When a complex number or vector points in the vertical direction, up or down, and you square it, then it rotates through a right angle and so the result is horizontal.
Vasco
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Post by Admin on Jul 17, 2022 0:38:09 GMT
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Post by mondo on Jul 17, 2022 2:58:47 GMT
Ok, you scaled it by a factor of .01 and it is visible now. Thank you.
So we figured out what was blocking me from understanding the work integral along a path $k$ from $1-i$ to $1+i$ for a vector field $H(z) = z$. The problem was I thought a vector field defined in chapter 10.1 is a vector originating at $z$ and ending at $H(z)$ (in regards to origin) while it is in regards to $z$. So that is why I couldn't get the length of $H(z) = z$ is not 0.
So now going back to the original problem of calculating work integral along $k$ - for $h(z) = z$ we just add up imaginary components of $z$ (because this is the tangential component of our vector from $z$ to $H(z)$. Since they will have opposite signs on each side of real axis (a symmetry) the integral will vanish.
Thank you Vasco!
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Post by Admin on Jul 17, 2022 8:03:32 GMT
Mondo
Well done for sticking with this! You got there in the end. You should now be able to study the rest of chapter 11. If you want any help don't hesitate to ask.
Vasco
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Post by mondo on Jul 18, 2022 6:32:01 GMT
Thank you Vasco.
I surely will. Since I read this book with breaks I plan to finish it with maybe 60% of understanding and then start over as there is a lot of material I already forgot/skipped.
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