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Post by mondo on Apr 10, 2022 22:54:25 GMT
Vasco,
then something is wrong. The geometric prove has only sense if $z$ is a vector from $C$ to the point $z$ on the circle with radius $\rho$. If $z$ is a vector from the origin as you said then you won't get $\frac{\pi}{2}$ between $z$ (a vector) and $z^2$...it is even visible from the Figure itself.
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Post by Admin on Apr 10, 2022 23:08:46 GMT
Mondo
There's nothing wrong with the proof. You are right that we need to show that the angle between $\rho$ and $z^2$ is a right angle. But we need to draw the dashed line as an extension of $z$ so that the red angle is $\theta$. then we show that the sum of the blue angle and the red angle is a right angle and so it then follows that the angle between $\rho$ and $z^2$ is also a right angle, which is what we want
Vasco
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Post by mondo on Apr 10, 2022 23:17:09 GMT
Mondo There's nothing wrong with the proof. You are right that we need to show that the angle between $\rho$ and $z^2$ is a right angle. But we need to draw the dashed line as an extension of $z$ so that the red angle is $\theta$. then we show that the sum of the blue angle and the red angle is a right angle and so it then follows that the angle between $\rho$ and $z^2$ is also a right angle, which is what we want Vasco I also think the proof is OK, the only thing we don't agree is this: "we need to draw the dashed line as an extension of $z$". I think it is just misunderstanding of what we define as $z$. If by $z$ you mean a vector from origin to the point $z$ on the circle then I don't see it. If on the other hand $z$ is really a radius $\rho$ from $C$ to the point $z$ then it is all ok. |
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Post by Admin on Apr 10, 2022 23:27:52 GMT
Mondo
$z$ is from the origin to the point $z$. We want to show that $z^2$ is tangent to the circle at the point $z$. This means that we must show that $z^2$ is at right angles to $\rho$, which is what the proof does. That doesn't mean that the dashed line should be in the direction of $\rho$. The dashed line is there to divide the angle of $z^2$ into two equal angles each equal to $\theta$
Vasco
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Post by mondo on Apr 10, 2022 23:52:32 GMT
I checked the algebraic method once more and I see that in your definition of $\rho$ you don't use $z$ and then it is OK, I agree $z$ can be thought of as a vector from $0$ to the point $z$ on our circle. I feel like I was confused because of the arrow at $z$ - you made it so it looks like $z$ is a vector from $C$. I think we are one the same page now. Thank you.
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Apr 11, 2022 5:18:13 GMT
Vasco, When you overlay known streamlines on the field, they pick out (touch) their own vectors. How did you write the expression for the vector field in EquationExplorer? Gary
Gary $z^2=(x+iy)^2=x^2-y^2+2xyi$ So real part is $x^2-y^2$ and imaginary part is $2xy$. The EquationExplorer uses the unit vectors $i$ and $j$ in the real and imaginary directions and so we write $z^2$ as $(x^2-y^2)i+2xyj$ Vasco Vasco,
Got it.
Gary
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