Write z=x+iy. Then |exp(-z)| = |exp(-x-iy)| = e^(-y). This is less than 1 (which is e^0) if and only if -y < 0, i.e., y > 0. Therefore, the answer is all complex numbers whose imaginary part is greater than zero.
Question 1(b)
iz -iz 2 iz -iz 2
2 2 (e - e ) (e + e )
sin z + cos z = (---------) + (----------)
( 2i ) ( 2 )
2iz 0 -2iz 2iz 0 -2iz
e - 2 e + e e + 2 e + e
= -------------------- + -------------------
-4 4
2iz -2iz 2iz -2iz
- e + 2 - e + e + 2 + e
= -------------------------------------
4
= 4/4 = 1
Question 2(a)
Log(1-i) = ln |1-i| + i Arg(1-i) = ln(sqrt(2)) + i(-pi/4) = 1/2 ln 2 - i pi/4.
log_(0,2pi)(1-i) = ln |1-i| + i arg_(0,2pi)(1-i) = 1/2 ln 2 + i 7pi/4.
Question 2(b)
1^i = exp(i log 1) = exp[i (ln 1 + i arg 1)] = exp[i (0 + i 2npi)] = e^(-2npi).
So, the values of 1^i are e^(-2npi) where n is an integer.
The principal value is exp[i (ln 1 + i Arg 1)] = exp[i(0 + i0)] = e^0 = 1.
Question 2(c)
The values of z^w are exp(w log z) = exp[w (Log z + 2npii] = exp[(w Log z) + (w 2 n pii)] = exp[(w Log z)] exp(2npii w) where n is an arbitrary integer.
In order for this to be single-valued, it must have the same value regardless of what n is. That means exp(2npii w) must have the same value it does when n=0, namely exp(0) = 1. So we need to have exp(2 npii w) = 1 for all integers n. As exp(2 n pii w) = exp(2 pii w)^n, this is the same as saying that exp(2 pii w) = 1.
The only complex numbers whose exponential is 1 are integral multiples of 2pii, so 2 pii w must be an integral multiple of 2 pii, which means that w must be an integer.
If you are uncomfortable with this line of reasoning, just write w = x+iy and observe that exp(2 pii w) = exp(2 pii x - 2 piy) = e^(-2 piy) (cos(2 pix) + i sin(2pix)). This is 1 if and only if y=0, cos(2pix) = 1, and sin(2pix) = 0, which implies that y is zero and x is an integer.
Question 3(a)
Parametrize C by z(t) = e^(it), 0 <=t <=2pi. Then z'(t) = ie^(it)dt, so
/ _ /2 pi it -it it
| |z| + z dz = | ( |e | + e ) i e dt
/C /0
/2 pi -it it
= | (1 + e ) i e dt
/0
/2 pi it
= | (i e + i) dt
/0
it |2 pi
= e + i t |
|0
= 1 + 2 pi i - (1 + 0)
= 2 pi i.
You can do this question even more simply by noting that |z|=1 on C and plugging in |z|=1 right from the start.
Question 3(b)
Recall that |e^(x+iy)| = e^x when x and y are real. Therefore, |exp(i Logz)| = |exp[i(ln |z| + i Arg z)]| = |exp( - Arg z + i ln |z|) | = e^(-Arg z) which lies between e^(-pi) and e^(pi) because -pi< Arg z <= pi by the definition of principal argument.
Therefore, |exp(i Logz)| <= e^(pi) for all z, and so
| / | pi pi | | exp (i Log z) dz | <= (e )(length of C) = 2 e . | /C |
Question 4(a)
C can be parametrized by x(t) = t, y(t) = t^2, 1 <= t <= 2. Then z(t) = t + t^2 i and z'(t) = 1 + 2ti, so
/ /2 2
| xyi dz = | t t i (1 + 2ti) dt
/C /1
/2 3 4
= | t i - 2 t dt
/1
4 5 |2
t 2t |
= -- i - --- |
4 5 |1
4 5 4 5
= 2 i / 4 - 2 2 / 5 - 1 i/ 4 + 2 1 /5
= 16 i/4 - 64/5 - i/4 + 2/5
= -62/5 + (15/4) i.
Question 4(b)
C starts at 1+1^2i = 1 + i and ends at 2 + 2^2 i = 2 + 4i. The integrand z + e^z is the derivative of z^2 + e^z. Therefore, by the fundamental theorem of calculus for contour integrals, the integral is
2 z |2+4i 2 2+4i 2 1+i
z + e | = (2+4i) + e - (1+i) - e
|1+i
2+4i 1+i
= 4 + 16i - 16 + e - 1 - 2i + 1 - e
2+4i 1+i
= -12 + 14i + e - e.
Question 5(a)
By Cauchy's integral formula,
1 / f(z) 1 1
f(0) = ------ | ------ dz = ------ i = ---- .
2 pi i /C z - 0 2 pi i 2 pi
Question 5(b)
The integrand is analytic everywhere except at z=-2i, which is outside the curve C. So the integrand is analytic everywhere on and inside the closed contour C, and therefore by the Cauchy-Goursat Theorem the integral is zero.