Remember that exp(x) is the usual e^x when x is real, exp(iy) = cos y + i sin y when y is real, and that exp(z+w) = exp(z)exp(w) for all complex numbers z and w.
(a) exp(2 +/- 3 pi i) = exp(2) exp( +/- 3 pi i) = e^2 (cos( +/- 3 pi) + i sin( +/- 3 pi)) = e^2 (-1 + 0 i) = -e^2.
(b) exp((2 + pi i)/4) = e^(2/4) (cos pi/4 + i sin pi/4) = sqrt(e) (1/sqrt(2) + i/sqrt(2)).
(c) exp(z + pi i) = exp(z) exp(pi i) = exp(z)(cos pi+ i sin pi) = exp(z)(-1 + 0i) = - exp(z).
Page 66 #3 (6th edition: page 68 #8)
First, some clarification: "ln" is the notation we use for the natural logarithm function of a positive real variable. It makes no sense to speak of "ln w" when w is a complex number (nor does it make sense when w is a negative real number, for that matter).
For complex variables, we let "log w" denote the set of all complex numbers z for which exp(z)=w. This is a multiple-valued function, unless we pick a particular branch.
These questions, however, come before we have even introduced the complex logarithm. You are simply asked to solve them using the definition of exp. It is wrong, for example, to try to tackle (a) by writing "z = ln(-2)"; the function ln is not defined at -2. And even if it were, it would only be giving you a single real-valued solution, not the set of all complex solutions as the question asked.
(a) We seek all values of z such that e^z = -2. Write z=x+iy; then e^z = e^x e^(iy), whereas -2 = 2 e^(ipi). These are equal if and only if e^x = 2 and y = pi+ 2n pi for some integer n (for recall that two complex numbers in polar form, re^(it) and Re^(iT), are equal if and only if r=R and t = T + 2n pi).
Since x is a real number, e^x=2 is equivalent to x = ln 2. Therefore, the values of z are x + iy = ln 2 + (2n+1)pi i where n is an integer.
(b) e^z = e^x e^(iy) and 1 + sqrt(3) i = 2e^(i pi/3). These are equal if and only if x = ln 2 and y = pi/3 + 2n pi. The values of z are therefore ln 2 + (2n + 1/3) pi i where n is an integer.
(c) If z = x+iy, then exp(2z-1) = exp(2x - 1 + 2iy) = e^(2x-1) e^(2iy), and 1 = 1 e^(i0). These are equal if and only if 2x - 1 = ln 1 = 0 and 2y = 0 + 2n pi. The first equation means x = 1/2; the second means y = n pi. The values of z are therefore (1/2) + n pi i where n is an integer.
Page 67 #5 (6th edition: page 68 #6)
Recall that |exp(w)| = exp(Re w) for all complex numbers w (the former is the complex exponential function; the latter is just the familiar real-variable exponential function). Therefore, |exp(z^2)| = exp(Re (z^2)). Now Re z^2 <= |z|^2, these are both real numbers, and exp is an increasing function on real numbers, so the fact that Re z^2 <= |z|^2 implies that exp(Re z^2) <= exp(|z|^2).
Page 67 #8 (6th edition: page 68 #11)
exp(x+iy) = e^x e^(iy) = e^x(cos y + i sin y) is a number whose modulus (distance to the origin) is e^x and whose argument (angle from the positive x-axis) is y.
(a) As x goes to - infinity , e^x -> 0, so exp(x+iy) ->0.
(b) As x remains fixed but y goes to infinity, exp(x+iy) will go round and round the circle of radius e^x, completing one full revolution each time y increases by 2pi.
Page 67 #14 (6th edition: page 68 #13)
They are harmonic, and V(x,y) is a harmonic conjugate of U(x,y), because the complex function F(x+iy) = U(x,y) + i V(x,y) is analytic.
The reason F is analytic is that F(x+iy) = e^(u(x,y))(cos v(x,y) + i sin v(x,y)) = exp(u(x,y) + i v(x,y)) = exp(f(x+iy)); in other words, F(z) = exp(f(z)). Since f and exp are analytic functions and the compositite of analytic functions is analytic, F is analytic.
Page 71 #11
Write sin(x+iy) = u(x,y) + i v(x,y) (if you wanted, you could find formulas for u(x,y) and v(x,y) by working from the definition of sin z, but you don't need to).
Then
_
sin z = sin (x - iy) = u(x,-y) + i v(x,-y).
Therefore, if we
write
_
sin z = U(x,y) + i V(x,y),
we have U(x,y) = u(x,-y)
and V(x,y) = v(x,-y), so the Cauchy-Riemann equations for
_
sin z
are U_x = V_y (i.e., u_x = -v_y),
and U_y = -V_x (i.e., -u_y = -v_x).
Because sin z is analytic we know that u_x = v_y, so the only way we can also have u_x = -v_y is if u_x = v_y=0. Similarly, we know that u_y = -v_x, so the only way we can also have -u_y = -v_x is if u_y = v_x = 0. Therefore, the only time the Cauchy-Riemann equations for
_
sin z
are satisfied is when u_x = u_y = v_x = v_y = 0, i.e.,
when sin' z = 0 (since sin' z = u_x + i v_x). This happens only when
cos z = 0, i.e., when z = pi/2 + 2n pi with n an integer. Therefore,
_
sin z
is differentiable only at these isolated points. Since
there is no point that has an entire disk around it where
_
sin z
is differentiable,
_
sin z
is not analytic anywhere.
The same argument works for cos z as well; in fact, it works for any non-constant analytic function.
Page 71 #12 (6th edition: page 72 #14)
First note that
_
exp(z) = exp(x - iy)
x
= e (cos(-y) + i sin(-y))
x
= e (cos y - i sin y)
___________________
x
= e (cos y + i sin y)
______
= exp(z)
for all complex numbers z.
_ _
_ exp(iiz) + exp(-iiz)
(a) cos(iz) = --------------------
2
_ _ _______ ______
exp(-z) + exp(z) exp(-z) + exp(z)
= ---------------- = ----------------
2 2
__________________ _______________________
exp(-z) + exp(z) exp(iiz) + exp(-iiz)
= ---------------- = --------------------
2 2
_______
= cos(iz).
_ _
_ exp(iiz) - exp(-iiz)
(b) sin(iz) = --------------------
2i
_ _ _______ ______
exp(-z) - exp(z) exp(-z) - exp(z)
= ---------------- = ----------------
2i ___
-2i
__________________ _______________________
exp(-z) - exp(z) exp(iiz) - exp(-iiz)
= ---------------- = - --------------------
-2i 2i
_______
= - sin(iz)
which equals _______ sin(iz)if and only if sin(iz) = 0, which happens if and only if iz = k pi where k is an integer, which happens if and only if z = (1/i) k pi= -i k pi= i n pi where n is an integer (the n here is the negative of the k in the previous formula).
Page 73 #5 (6th edition: page 74 #7)
sinh(z + pi i) = ( exp(z + pi i) - exp(-z - pi i))/2 = ( exp(z) exp(pi i) - exp(-z) exp(-pi i))/2 = (-exp(z) + exp(-z))/2 = - sinhz. The same argument works for cosh.
tanh(z + pi i) = sinh(z + pi i) / cosh(z + pi i) = ( - sinh z) / (-cosh z) = (sinh z)/(cosh z) = tanh z.
Page 78 #1
(a) Log(-ei) = ln |-ei| + i Arg(-ei) = ln e + i (-pi/2) = 1 - pi i/2.
(b) Log(1-i) = ln |1-i| + i Arg(1-i) = ln sqrt(2) + i(-pi/4) = (ln 2)/2 - i pi/4.
Page 78 #2
(a) log e = ln |e| + i arge = ln e + i(0 + 2npi) = 1 + 2npi i where n is an integer.
(b) log i = ln |i| + i argi = ln 1 + i(pi/2 + 2npi) = 0 + (1/2 + 2n)pi i where n is an integer.
(c) log(-1 + sqrt(3) i) = ln |-1 + sqrt(3) i| + i arg(-1 + sqrt(3) i) = ln 2 + i(2pi/3 + 2npi) = ln 2 + 2(1/3 + n)pi i where n is an integer.
Page 78 #3
2 Log(1+i) = 2(ln |1+i| + i Arg(1+i)) = 2(ln sqrt(2) + i pi/4) = 2((ln 2)/2 + i pi/4) = ln 2 + i pi/2.
Similarly, 2Log(-1+i) = 2((ln 2)/2 + i 3pi/4) = ln 2 + i 3pi/2.
(1+i)^2 = 2i and (-1+i)^2 = -2i. Therefore, Log(1+i)^2 = Log(2i) = ln|2i| + i Arg(2i) = ln 2 + i pi/2 which is the same as 2 Log(1+i). However, Log(-1+i)^2 = Log(-2i) = ln |-2i| + i Arg(-2i) = ln 2 - i pi/2 which is different from 2 Log(-1+i).
Page 78 #4
(a) The value of argi that lies in the given range is pi/2, and the value of arg(i^2) = arg(-1) that lies in the given range is pi.
Therefore, with this branch of the logarithm, log(i^2) = ln |i^2| + pi i = 0 + pi i, and 2 log i = 2(ln |i| + i pi/2) = 0 + pi i = log(i^2).
(b) The value of argi that lies in the given range is 5pi/2, and the value of arg(i^2) = arg(-1) that lies in the given range is pi.
Therefore, with this branch of the logarithm, log(i^2) = ln |i^2| + pi i = 0 + pi i, and 2 log i = 2(ln |i| + i 5pi/2) = 0 + 5 pi i which is different from log(i^2).
Page 78 #5
(a) Recall that the two values of i^(1/2) are +/- (1+i)/sqrt(2). (Refer back to earlier assignments if you're rusty on calculating complex roots). Now log (1+i)/sqrt(2) = ln|(1+i)/sqrt(2)| + i arg (1+i)/sqrt(2) = ln 1 + i pi/4 + 2npi i = (2n + 1/4)pi i where n is an integer. Similarly, log (-(1+i)/sqrt(2)) = -i 3pi/4 + 2kpi i = (2n - 3/4)pi i where n is an integer. Let's rewrite this as ((2n-1) + 1/4)pi i.
The set of values for log(i^(1/2)) is the union of the above two sets of values. Therefore, it consists of the numbers (2n + 1/4)pi i as n ranges over all the integers, along with the numbers ((2n-1) + 1/4)pi i as n ranges over all the integers.
The former set covers all the values (k+1/4)pi i where k ranges over the even integers, and the latter set covers these values as k ranges over the odd integers. Therefore, the set of values for log(i^(1/2)) consists of the numbers (k + 1/4)pi i as k ranges over all the integers. Renaming k to n gives the answer in the book.
The set (1/2) log i = (1/2)(ln|i| + i pi/2 + 2n pi i) = i pi/4 + n pi i = (n+1/4)pi i, so this is the same set as above, proving log(i^(1/2)) = (1/2)log(i).
(b) log(i^2) = log(-1) = ln|-1| + i pi + i 2npi = (2n+1)pi i where n is an arbitrary integer. However, 2 log i = 2( ln|i| + i pi/2 + 2npi i) = i pi + i 4npi = (4n+1)pi i which is not the same set as log(i^2). (It is a proper subset, consisting of those values (2n+1)pi i where n is even).
Page 79 #10
Log(z1 z2) = ln|z1 z2| + i Arg(z1 z2) = ln|z1| + ln|z2| + i Arg(z1 z2), whereas Log z1 + Log z2 = ln|z1| + ln|z2| + i Arg z1 + i Arg z2.
Therefore, what we need to show is that Arg(z1 z2) = Arg z1 + Arg z2 + 2Npi where N is 0, 1, or -1.
Remember that Arg z1 + Arg z2 is always a value of arg(z1 z2); just not necessarily the principal one. Therefore, it differs from the principal argument by a multiple of 2pi. This proves that Arg(z1 z2) = Arg z1 + Arg z2 + 2Npi where N is an integer. It remains to show that N is 0, 1, or -1.
To do this, write 2Npi= Arg(z1 z2) - Arg z1 - Arg z2. All terms on the RHS, being principal arguments, are between -pi and pi, so the RHS is >= -pi - pi - pi = -3pi and is <= pi + pi + pi= 3pi. Therefore, -3pi <= 2Npi <= 3pi, so -3/2 <= N <= 3/2. Since N is an integer, this means N is -1, 0, or 1.
Alternative solution for the last part
Another way to think of this is to note that Arg z1 + Arg z2 lies between -pi-pi=-2pi and pi+pi=2pi. That gives three cases:
Case 1: it is between -2pi and -pi, in which case adding 2pi to it will bring it into the correct range for a principal argument, so Arg(z1 z2) = Arg(z1) + Arg(z2) + 2pi.
Case 2: it is between -pi and pi, in which case it is the principal argument, so Arg(z1 z2) = Arg(z1) + Arg(z2).
Case 3: it is between pi and 2pi, in which case subtracting 2pi from it will bring it into the correct range for a principal argument, so Arg(z1 z2) = Arg(z1) + Arg(z2) - 2pi.
Page 79 #14 (6th edition: #13)
(a) Recall that Log is analytic everywhere except at the origin and negative real axis, that is, except at numbers of the form t + 0i (t real) where t <= 0. The function f(z)=z-i is analytic everywhere, and the composition of analytic functions is analytic, so Log(z-i) is analytic except when z-i = t + 0i with t <= 0. If we write z=x+iy, then z-i = x + i (y-1), so Log(z-i) is analytic everywhere except where y=1 and x <= 0.
(b) This is analytic everywhere except where the denominator z^2+i is zero (which happens at the two points z = +/- (1-i)/sqrt(2)) and except where z+4 is a non-positive real number, which happens when z is a real number <= -4.