
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.

Performing operations on numbered pairs
Z1 = (1,1)
Z2 = (1,1)
Z3 = (34)
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)

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.

Matrices : GaussJordan Elimination
:
x1+x2+x3= 10
y1+y2+y3=10
z1+z2+z3=10
x1+y1+z1x2y2z2=0
x2+y2+z2x3y3z3=0
x1+1/2y1x21/2y2=0
x2+1/2y2x31/2y3=0
This gives us the following augmented matrix:
Note that the order of equations are not

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.

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).

Complex Polynomial Proof : Factoring
If we divide p(z) by (zz1), it is clear that the quotient is a polynomial of degree (n1) and the highest power of z is a0 z ^(n1)
Therefore p(z) = (zz1) (a0 z^n1+......)
Let the second factor be Q1(z)
Therefore p(z) = (zz1)Q1(z).

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).

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) / (