SOLUTIONS - San Francisco State University
cut from the solid cylinder x2 + y2 4 by the planes z= 0 and z= 1 Again, we can apply Gauss’s theorem Note that y 2+ 4z = 16 4z= 2 y 4y 16 = 8 z Since 2z= y
Complex Variables 05-3, Exam 1 Solutions, 7/14/5 Question 1
jzj = 3; jz 4j = 5; zz = 9; (z 4)(z 4) = 25; zz = 9; zz 4z 4z +16 = 25; x 2+y = 9; z +z = 0; x = 0; y2 = 9; ; (x;y) = (0; 3); z = 3i: A parameterization for the boundary arcs is then: { z = 3e it; ˇ 2 t ˇ 2 { z = 4+5eis; a s a Here the endpoint parameter value a is given by: 4+5eia = 3i; 5eia = 3i 4; eia = 4+3i 5 = cos(a)+isin(a); cos(a
53 sections 202/204 Quiz 7 Solutions
the paraboloid and the plane x= 4, projected to the yzplane This intersection is the circle y 2+ z = 1 (set x= 4 in the equation x= 4y2 + 4z) ZZZ E xdV = ZZ y 2+z =1 Z 4 4y2+4z2 xdxdydz = ZZ y 2+z =1 1 2 (16 216(y2 + z)2)dydz = 8 Z 2ˇ 0 Z 1 0 (1 r4)rdrd changing to polar for the yzplane = 16ˇ 1 2 1 6 = 16ˇ 3
10 Cosets and the Theorem of Lagrange
x10 Cosets and the Theorem of Lagrange Homework: 1-7, 12-16, 26, 28-33 1 Find all cosets of the subgroup 4Zof Z 4Z= f ; 8; 4;0;4;8;g 1 + 4Z= f ; 7; 3;1;5;9;g
Ma 416: Complex Variables Solutions to Homework Assignment 8
in the annulus 1/4 < zfor the function f(z) = z−2(4z −1)−1 Solution: In the punctured disk 0 < z< 1/4, write f(z) = 1 z2(4z −1) = 1 z2 X ∞ n=0 (−4 n)zn = −16 X n=−2 4nz It is clear that the radius of convergence is 1/4 and that there is a pole of order 2 at z = 0 In the annulus z> 1/4, write f(z) = 1 z2(4z −1) = 1/z3
6 2 6 = 0 - UMD
To find the critical points we need to solve 6x2 − 6x = 0, 5y4 − 20 + 3y 2z = 0, and 2y3z = 0 From the first equation we get x = 0,1 From the third equation we get y = 0 or z = 0 But y = 0 violates the second equation so we must have z = 0 Plugging z = 0 into the second equation gives y = ± √ 2 Note that when z = 0 the Hessian is
44 Systems of Equations - Three Variables
13) 2x + y − 3z =0 x − 4y+ z =0 4x + 16y+4z =0 15) 3x +2y+2z =3 x +2y − z =5 2x − 4y+ z =0 17) x − 2y+3z =4 2x − y+ z = − 1 4x + y+ z =1 19) x − y+2z =0 x − 2y+3z = − 1 2x − 2y+ z = − 3 21) 4x − 3y+2z = 40 5x +9y − 7z = 47 9x +8y− 3z = 97 2) 2x +3y= z − 1 3x =8z − 1 5y+7z = − 1 4) x + y+ z =2 6x − 4y+5z
1 Inverse Z-transform - Partial Fraction
z 2 Digital Control 5 Kannan M Moudgalya, Autumn 2007 6 Important Result from Di erentiation 6 n=0 anz n; az 1
Functions of a ComplexVariable(S1)
z2 +4z +1 3 C1 z z+ z+ = − 2 + •By residue theorem I = 2πi [Resz=z+f]= 2π √ 3 I1 = Z 2π 0 dθe−cosθ cos(θ +sinθ), I 2 = Z 2π 0 dθe−cos
Homework 2 Model Solution - Han-Bom Moon
0 4 2 = h 4; 2;4i P 1: a plane containing L 1 and having v 1 v 2 as a normal vector 4(x 1) 2(y 2)+4(z 6) = 0 ) 4x+4 2y+4+4z 24 = 0) 4x 2y+4z 16 = 0 P 2: a plane containing L 2 and having v 1 v 2 as a normal vector 4(x 3) 2(y 4)+4(z 0) = 0 ) 4x+12 2y+8+4z= 0) 4x 2y+4z+20 = 0 distance between L 1 and L 2 = distance between P 1 and P 2 = distance
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