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Impedances
19640 Impedances for Imaginary and Real Parts Let Z1 = (-1,-1) and Z2 = (1,1). Perform the operation Z1 and Z2. Use number pairs.
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Performing operations on numbered pairs
Z1 = (-1,-1)
Z2 = (1,1)
Z3 = (3-4)
Z4 = (-sqrt(3),1)
Z4* = (-sqrt(3),-1)
Z2+Z3 = (4,-3)
Z1(Z2+Z3) = (-1*4 - (-1)*(-3), (-1)*(-3) +(-1)*(4))
=> Z1(Z2+Z3) = (-7,-1)
Z1(Z2+Z3)Z4* = (-7,-1)(-sqrt(3),-1)
=> Z1(Z2+Z3)Z4* =((-7)(-sqrt(3)) - (-1)
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Microeconomics Questions: Production and Game Theory
Before we proceed let's compute d f(z1,z2) / dz1 = 1/ (z1^0.5) and d f(z1,z2) / dz2 = 1/ (z2^0.5). Hint: we need to use the same rule as in part 1 (i).
So, let's find d Π(z1,z2) / d z1 and d Π(z1,z2) / d z2.
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Matrices : Gauss-Jordan Elimination
:
x1+x2+x3= 10
y1+y2+y3=10
z1+z2+z3=10
x1+y1+z1-x2-y2-z2=0
x2+y2+z2-x3-y3-z3=0
x1+1/2y1-x2-1/2y2=0
x2+1/2y2-x3-1/2y3=0
This gives us the following augmented matrix:
Note that the order of equations are not
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Fractional Transformations, Cross Ratios and Conformal Mapping
99894 Fractional Transformations, Cross Ratios and Conformal Mapping 1. a) Let z1,z2,z3,z4 lie on a circle.
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Equivalence relation on set of ordered pairs
So let (z1, z2), (z3, z4), and (z5, z6) be ordered pairs of positive integers, and suppose that (z1, z2) R (z3, z4) and (z3, z4) R (z5, z6).
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Complex Polynomial Proof : Factoring
If we divide p(z) by (z-z1), it is clear that the quotient is a polynomial of degree (n-1) and the highest power of z is a0 z ^(n-1)
Therefore p(z) = (z-z1) (a0 z^n-1+......)
Let the second factor be Q1(z)
Therefore p(z) = (z-z1)Q1(z).
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Complex Rotations
Let d(z1, z2) denote the distance between two complex numbers, z1 and z2. By problem 1, we see that
d(z1, z2) = |z1 - z2|.
Thus we have:
d(uv, uw) = |uv - uw| = |u(v - w)|.
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Two Inequalities Involving Complex Numbers
that sqrt(2) * |z| >= |Re z| + |Im z|
(suggestion: reduce this inequality to (|x| - |y|)^2 +. 0 1) We have
Re(z1 + z2) <= |z1 + z2| <= |z1| + |z2|
and
|z3 + z4| >= ||z3| - |z4||,
whence
Re(z1 + z2) / |z3 + z4| <= (|z1| + |z2|) / (||