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MAT334--Lectures & Home Assignments / Mobius tranformation
« on: December 01, 2018, 05:56:54 PM »
What is the formula or way for finding the rotation angle (arg(𝑓′(𝑧))) for the derivative of Mobius transformation?
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MAT244--Lectures & Home Assignments / Final Review 9.2 and 9.3
« on: November 27, 2018, 09:48:48 PM »
I would appreciate if anyone can share the technique for drawing phase planes by hand. I get the idea of linearization and finding the equilibrium points and classifying those but have difficulty in sketching the phase planes. I am not sure about how exactly to draw it by hand.
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MAT334--Lectures & Home Assignments / Section 2.6
« on: November 21, 2018, 10:29:04 AM »
Does anyone know how to approach questions 4 to 8 in 2.6? It says to follow example 3- 6 but I tried and I am stuck. Any help would be appreciated
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MAT334--Lectures & Home Assignments / Help with attempting questions 17 -20 1.6
« on: October 09, 2018, 05:01:52 PM »
I am not sure on how to approach the questions 17-20 from 1.6.
For 17 I started off using the green's theorem
$$ \int_{\gamma} (Pdx + Qdy) = \iint_{\omega} \Bigl[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \Bigr]dxdy
$$
Since $Pdx + Qdy$ is exact differential $P = \frac{\partial g}{\partial x}$ and $Q = \frac{\partial g}{\partial y}$
$$\frac{\partial P}{\partial y} = \frac{\partial ^2 g}{\partial x \partial y} \\
\frac{\partial Q}{\partial x} = \frac{\partial ^2 g}{\partial x \partial y}$$
So $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$
Hence $$\int_{\gamma} (Pdx + Qdy) = \iint_{\omega} \Bigl[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigr] dxdy= 0$$
Not sure if this is correct and how to proceed with 18-20
For 17 I started off using the green's theorem
$$ \int_{\gamma} (Pdx + Qdy) = \iint_{\omega} \Bigl[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \Bigr]dxdy
$$
Since $Pdx + Qdy$ is exact differential $P = \frac{\partial g}{\partial x}$ and $Q = \frac{\partial g}{\partial y}$
$$\frac{\partial P}{\partial y} = \frac{\partial ^2 g}{\partial x \partial y} \\
\frac{\partial Q}{\partial x} = \frac{\partial ^2 g}{\partial x \partial y}$$
So $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0$
Hence $$\int_{\gamma} (Pdx + Qdy) = \iint_{\omega} \Bigl[\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigr] dxdy= 0$$
Not sure if this is correct and how to proceed with 18-20
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MAT334--Lectures & Home Assignments / Section 1.4 Question 1
« on: October 01, 2018, 08:02:20 PM »
lim n -> infinity zn = ((1 +i)/(sqrt(3))n.
I used polar coordinates to solving giving r = sqrt(2/3)
xn =sqrt(2/3) cos(n * pi /4)
yn =sqrt(2/3) sin(n * pi /4)
zn = sqrt(2/3) cos(n * pi /4) + isqrt(2/3) sin(n * pi /4)
From this how can show that it converges to 0?
I used polar coordinates to solving giving r = sqrt(2/3)
xn =sqrt(2/3) cos(n * pi /4)
yn =sqrt(2/3) sin(n * pi /4)
zn = sqrt(2/3) cos(n * pi /4) + isqrt(2/3) sin(n * pi /4)
From this how can show that it converges to 0?
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MAT334--Lectures & Home Assignments / To show z1z2 = 0 implies z1 = 0 or z2 = 0
« on: September 24, 2018, 03:27:11 PM »
Will a prove which uses rectangular coordinates definition of z1 and z2 would be fine?
Using z1 = x1 +iy1 and z2 = x2 + iy2 and using some algebra to prove z1z2 = 0 implies z1 = 0 or z2 = 0
Barely comprehensible. Next such post will be deleted. V.I.
Using z1 = x1 +iy1 and z2 = x2 + iy2 and using some algebra to prove z1z2 = 0 implies z1 = 0 or z2 = 0
Barely comprehensible. Next such post will be deleted. V.I.
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MAT334--Lectures & Home Assignments / Section 1.1 question 3 part f and g
« on: September 17, 2018, 07:18:12 PM »
$\renewcommand{\Re}{\operatorname{Re}}$
How is the sketch for locus
(f) $\,\Re[(1-i) \bar{z} ] = 0$
(g) $\Re[z/(1+i)] = 0$
How is the sketch for locus
(f) $\,\Re[(1-i) \bar{z} ] = 0$
(g) $\Re[z/(1+i)] = 0$
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