Download Home Exercise

BGU Physics Dept. Introduction to Mathematical Methods in Physics
Home Exercise
Complex Numbers
1. Solve the following equations:
(a) z 2 − 5z + 10 = 0
(b) 4z 3 − 6z 2 + 3z = 0
solution
(a) z1 =
5
2
+ 2i, z2 = z1 =
(b) z1 = 0, z2 =
3
4
√
+i
12
8 ,
5
2
− 2i
z3 =
3
4
√
−i
12
8
2. For z = 4i and w = 2 − 3i, calculate
(a) z + w
(b) z − w
(c) z · w
(d)
z
w
(e) z ∗ w − zw∗
(f) w2
(g) ln(w)
√
(h) 1 + z + w
solution
(a) z + w = 2 + i
(b) z − w = 2 − 7i
(c) z · w = 12 + 8i
(d)
z
w
=
12
13
+
8
13 i
(e) z ∗ w − zw∗ = −24
(f) w2 = 5 − 12i
(g) ln(w) = ln(13)
+ i tan−1 − 32 + 2πn
2
√
√
(h) 1 + z + w = ± 4 10 (cos(θ/2) + i sin(θ/2)) θ = tan−1 (1/3)
3. Calculate
(a) Re(e4+3i )
(b) Im(22i+3 )
solution
(a) e4
1
(b) 22i+3 = e(3+2i) ln 2 = e3 ln 2 ei2 ln 2 ⇒ Im(2i+3 ) = e2 ln 2 = eln 4 = 4
4. (a) Show that cos(4θ) = 8 cos4 (θ) − 8 cos2 (θ) + 1
q √
(b) Use the above result to prove that cos π8 = 2+4 2
solution
(a)
cos(4θ) + i sin(4θ) = (cos θ + i sin θ)4
= cos4 θ + 4i cos3 θ sin θ − 6 cos2 θ sin2 θ − 4i cos θ sin3 θ + sin4 θ
cos(4θ) = Re(cos(4θ) + i sin(4θ)) = cos4 θ − 6 cos2 θ sin2 θ + sin4 θ
= cos4 θ − 6 cos2 θ(1 − cos2 θ) + (1 − cos2 θ)2 = 8 cos4 θ − 8 cos2 θ + 1
(b) For θ =
π
8
we have
π π π = 0 = 8 cos4
− 8 cos2
+1
cos
2
8
8
Let x = cos2 π8 to obtain:
8x2 − 8x + 1 = 0
s
√
π 2+ 2
x = cos
=
8
4
5. Using euler theorem and the multiplication eiθ eiφ , find trigonometric identities for:
(a) sin2 (θ)
(b) sin(θ + φ)
(c) cos(θ + φ)
solution
(a)
2
sin (θ) =
eiθ − e−iθ
2i
2
=−
1 − cos(2θ)
1 2iθ
e + e−2iθ − 2 =
4
2
(b) + (c)
cos(θ + φ) + i sin(θ + φ) = ei(θ+φ) = eiθ eiφ = (cos θ + i sin θ) (cos φ + i sin φ)
= cos θ cos φ − sin θ sin φ + i (cos θ sin φ + sin θ cos φ)
cos(θ + φ) = cos θ cos φ − sin θ sin φ
sin(θ + φ) = cos θ sin φ + sin θ cos φ
2
if any typos or corrections found please let me know: Ben Yellin [email protected]
3