Math 250: Solutions of Homework II
1. Proof. Let z be a point on the tangent line, then we have
( B)(B C) + (z B)(B C) = 0.
z
So
(B C)z + (B C) = 2B B B C BC
z
= B B + (B B B C BC + C C C C
= B B + (B C)(B C) C C
= B B C C + r2
2. Solution. The three tra
Math 250: Solutions of Homework I
s 1
t
1. Solution. A = ( s+t , 0), B = ( 2 , 2 ) and C = ( 2 , 1 ). Also, the coordinates for
2
2
centroid G is ( s+t , 1 ).
3
3
2. Solution. Use the facts that ADBC and D is in the line BC we have D = (0, 0),
2
2
2 st
2
Math 250: Solutions of Homework V
1. Proof.
ax + ay
x+y
z =
(x y) z =
xy z =
1+ a
a + xy
x (y z) = x
ax+ay
a+xy
1+
+z
ax+ay
z
a+xy
=
ax + ay + az + xyz
.
a + xy + yz + zx
a
x + ay+az
ay + az
ax + ay + az + xyz
y+z
a+yz
=x
=
.
ay+az =
yz
x a+yz
1+ a
a + y
Math 250: Solutions of Homework VI
1. Proof. By carefully computation, we have a = r21 q23 r3 r23 q21 r1 , b = r23 q21 q3 r1
r21 q23 q1 r3 , c = r21 q23 r23 q21 and d = r23 q21 q3 r21 q23 q1 , where rij and qij mean ri rj
and qi qj respectively. Hence a,
Math 250: Solutions of Homework III
1. Proof. It is similar to the proof of Proposition 3.2 , page 7 in the textbook of week 3.
1
1
1
2. Proof. Since ( 3 )2 + ( 3 )2 + ( 3 )2 = 1, we know that the 8 points are on the
Riemann sphere. To see the image of th
Math 250: Solutions of Homework X
1. Solution. The dual coordinate for the line through the two points is [(1, 1, 0)
(1, 2, 0)] = [0, 0, 1]. Hence the line through these two point is z = 0.
2. Solution. The line equation is (x, y, z) = t(3, 1, 2).
3. Sol
Math 250: Solutions of Homework VIII
1. Solution.
Solve f (z) = z, we get that the xed points of f is z =
2
3
3
i.
2
det(f ) = 1. Hence the 2 2 matrix represents f is
1 3
1 2
.
tr 2 (f ) = (1 2)2 = 1.
1
Solve 2 + (2 tr 2 (f ) + 1 = 0, then the multipl
Math 250: Solutions of Homework VII
1. Solution.
(1) z = 1
3i.
(2) z = 0 and z = 5.
(3) z = 0 and z = .
(4) z =
7 17
.
2
2. Solution.
(1) det(f ) = 1, hence the matrix is
1 2i
i 3
2
6i
1
6i
(2) det(f ) = 6, hence the matrix is
5
0
(3) det(f ) = 5, hence
Math 250: Solutions of Homework IV
1. Solution. For a dilation, the xed points are 0 and . For a translation, the xed
point is . For a general ane transformation z az + b, the xed point are and
b
if a = 1. For the inversion, the xed points are 1.
1a
2. So