---------------
#  Aufgabe 1  #
---------------

n = 100000
phi = seq(from=0,to=2*pi,length=n)

# 1a)

z0 = 1
r = 1
z = z0 + r*exp(1i*phi)
dz = diff(z)
z = z[-1]
fz = 1/( (z-1)^2 * (z^2+4) )

intf = sum( fz*dz )
intf                   # -0.0000158-0.5026548i
-4*pi*1i/25            #          0-0.5026548i

# 1b)

z0 = 0
r = 4
z = z0 + r*exp(1i*phi)
dz = diff(z)
z = z[-1]
fz = 1/( (z-1)^2 * (z^2+4) )

intf = sum( fz*dz )
intf                   # 4.5121e-20 - 5.481574e-18i
                       # exact result: 0, das passt.


---------------
#  Aufgabe 2  #
---------------

n = 100000
xmax = 100
x = seq(from=-xmax,to=xmax,length=n)
dx = diff(x)
x = x[-1]

# 2a)

fx = 1/(1+x^2)^2
intf = sum( fx*dx )
intf                   # 1.570796
pi/2                   # 1.570796

# 2b)

fx = 1/(1+x^4)
intf = sum( fx*dx )
intf                   # 2.221441
pi/sqrt(2)             # 2.221441



---------------
#  Aufgabe 4  #
---------------

# Im Inneren von C_epsilon(0) darf sich nur 
# die Polstelle bei z=0 befinden, keine 
# anderen Pole wie z=-1 oder z=pi. Wir 
# waehlen 

eps = 0.1
n = 100000
phi = seq(from=0,to=2*pi,length=n)
z = eps*exp(1i*phi)
dz = diff(z)
z = z[-1]

fa = cos(z)/z^3
fb = (z^2+4*z+5)/(z^2+z)
fc = exp(z)/z^2
fd = (exp(2*z)-1)/(sin(z))^2

Res_a = 1/(2*pi*1i)*sum( fa*dz )
Res_b = 1/(2*pi*1i)*sum( fb*dz )
Res_c = 1/(2*pi*1i)*sum( fc*dz )
Res_d = 1/(2*pi*1i)*sum( fd*dz )

Res_a       # exact: -0.5
Res_b       # exact: 5
Res_c       # exact: 1
Res_d       # exact: 2