MATH 300 Problem Set 4 Solutions In this problem set logτ z refers ...

MATH 300 Problem Set 4 Solutions
In this problem set log τ z refers to the branch of logz for which τ < Im log τ z ≤ τ +2π.
1. Factor the following polynomials into a product of linear terms:
(a) z 2 +3z −17i
(b) z 3 +z 2 −z −1
(c) z 5 −1
Solution. (a) By the high school quadratic formula, the two roots of z2 +3z −17i are −3±√9+68i 2
we express 9+68i = reiθ , where r = √ 92 +682 = √ −1 68 π
4705 and θ = tan 9 (with 0 < θ < 2
two square roots of 9+68i are ± √ reiθ/2 and the two roots of z2 +3z −17i are −3 1
2 ± 2
z 2 +3z −17i = z + 3 √ 1 iθ/2
2 + 2 re z + 3 √ 1 iθ/2
2 − 2 re
(b) Subbing in z = 1 shows that 1 is a root of z 3 +z 2 −z −1. Factoring
z 3 +z 2 −z −1 = (z −1) z 2 +2z +1 = (z −1)(z +1)(z +1)
(b) The roots of z 5 −1 are the five fifth roots of 1, namely 1, e i2
5 π , e i4
5 π , e i6
5 π and e i8
5 π . So
z 5 −1 = (z −1) z −e i2
5 π z −e i4
5 π z −e i6
5 π z −e i8
5 π
2. Find the poles of the following rational functions
(a) z−1
z 2 +1
(b) z 2
z 4 +4
(c) z+1
z 2 +z+1
Solution. (a) The poles are at the zeros of the denominator z 2 +1. That is, at z = ±i.
. If
), then the
√ iθ/2 re . Hence
(b) The poles are at the zeros of the denominator z4 +4. That is, at the four fourth roots of −4. That
is, at z = √ 2eiπ 4, √ 2ei3π 4 , √ 2ei5π 4 and √ 7π i 2e 4 .
(c) The poles are at the zeros of the denominator z2 +z+1. By the high school quadratic formula, the
two roots are −1±√ 1−4
2 = − 1
2 ± √ 3
2 i.
3. Show that the formula e iz = cosz +isinz holds for all complex numbers z.
Solution. By definition
4. Verify the identity
cosz +isinz = eiz +e −iz
2 +i eiz −e −iz
2i = eiz +e −iz
2 + eiz −e −iz
2 = e iz
sinz1 −sinz2 = 2cos z1+z2
2
z1−z2
sin 2
and use it to show that sinz1 = sinz2 if and only if z2 = z1 + 2kπ or z2 = −z1 + (2k + 1)π for some
integer k.
Solution. Multuplying out
2cos z1+z2
2
e z1−z2
sin 2 = 2 iz 1 +z2 2 +e−iz 1 +z2 2
2
= eiz1 +e −iz2 −e iz2 −e −iz1
2i
= sinz1 −sinz2
1
eiz 1−z2 2 −e−iz 1−z2 2
2i
= eiz1 −e −iz1
2i
− eiz2 −e −iz2
2i

as desired. Consequently,
sinz1 = sinz2 ⇐⇒ sinz1 −sinz2 = 0 ⇐⇒ cos z1+z2 z1−z2
2 sin 2 = 0
⇐⇒ cos
z1+z2 z2−z1
2 = 0 or sin 2 = 0
⇐⇒ z1+z2 1 z2−z1
2 = 2 (2k +1)π or 2 = kπ for some k ∈ Z
⇐⇒ z2 = −z1 +(2k +1)π or z2 = z1 +2kπ for some k ∈ Z
Here we have used that
cosz = 0 ⇐⇒ e iz +e −iz = 0 ⇐⇒ e iz = −e −iz ⇐⇒ e 2iz = −1 ⇐⇒ z = 1
2 (2k +1)π for some k ∈ Z
sinz = 0 ⇐⇒ e iz −e −iz = 0 ⇐⇒ e iz = e −iz ⇐⇒ e 2iz = 1 ⇐⇒ z = kπ for some k ∈ Z
5. Show that the mapping w = sinz is one–to–one in the semi–infinite strip
S1 = x+iy −π < x < π, y > 0
That is, show that if z1 and z2 are in S1 and sinz1 = sinz2, then z1 = z2. Find the image of this strip.
Solution. Let z1 = x1 +iy1 and z2 = x2 +iy2 both be in the strip S1. By the last question,
sinz1 = sinz2 ⇐⇒ z2 = −z1 +(2k +1)π or z2 = z1 +2kπ for some k ∈ Z
⇐⇒ x2 = −x1 +(2k +1)π, y2 = −y1 or x2 = x1 +2kπ,y2 = y1 for some k ∈ Z
Since both x1 + iy1 and x2 + iy2 lie in S1, we necessarily have both y1 > 0 and y2 > 0. So “x2 =
−x1 +(2k+1)π, y2 = −y1 for some k ∈ Z” cannot happen. We also necessarily have the −π < x1 < π
and −π < x2 < π and hence |x1 −x2| < 2π. So “x2 = x1 +2kπ, y2 = y1 for some k ∈ Z” can happen
only for k = 0, in which case z1 = z2. This completes the proof that sinz is one–to–one in S1.
We now determine the image of S1. For z = x+iy,
sinz = sin(x+iy) = eix−y −e −ix+y
2i
= ey +e −y
2 sinx+i ey −e −y
2 cosx
= 1
2i
e −y (cosx+isinx)−e y (cosx−isinx)
For each fixed y > 0, as x runs from −π to π, the point sin(x+iy) = ey +e −y
2 sinx+i ey −e −y
2 cosx runs
around an ellipse (starting just to the left of the negative imaginary axis near −i ey −e −y
2 and ending
just to the right of the negative imaginary axis near −i ey −e −y
2 ) with semimajor axis ey +e −y
2 > 1 and
semiminor axis ey −e −y
2 > 0. When y > 0 is very close to zero, the semimajor axis is just a bit bigger
than 1 and the semiminor axis is just a bit bigger than 0, as for the smallest ellipse in the figure below.
As y increases both semiaxes increase in size and both tend to infinity as y tends to infinity. For y large,
the ellipse is almost circular, as for the largest ellipse in the figure below. The set of points covered is
−π < x < π
y>0
sin(x+iy) = ey +e −y
2 sinx+i ey −e −y
2 cosx
which is the union of all of the ellipses, excluding the intersection of the ellipse with the negative
imaginary axis. This is the entire complex plane, except for the negative imaginary axis and the line
segment y = 0, −1 < x < 1, as sketched in the figure
2

6. Find all values of the following:
(a) Log(1−i)
(b) log(1+i)
(c) log 3π (−1+ √ 3i)
Solution. (a) In polar coordinates 1−i = √ 2e−iπ 4 = e 1
2 ln2−iπ 4. So log(1−i) = 1
2 ln2−iπ 4 +2kπi with
k running over Z. The principal value of the logarithm chooses the k for which the imaginary part is
between −π and π, namely k = 0. So Log(1−i) = 1
2 ln2−iπ 4 .
(b) In polar coordinates 1+i = √ 2eiπ 4 = e 1 π
2 ln2+i 4. So log(1+i) = 1
2 ln2+iπ 4 +2kπi with k running
over Z.
(b) In polar coordinates −1+ √ 3i = 2ei2π 2π
3 ln2+i = e 3 . So log(−1+ √ 3i) = ln2 + i2π 3 + 2kπi with k
running over Z. log 3π chooses the k for which the imaginary part is between 3π and 5π, namely k = 2.
So log 3π (−1+ √ 3i) = ln2+i 2π
3
+4πi = ln2+i14π
3 .
7. Prove that Loge z = z if and only if −π < Imz ≤ π.
Solution. We have loge z = z + 2kπi with k ∈ Z. For the principal branch we choose k ∈ Z so that
−π < ImLoge z ≤ π. Since Re(z +2kπi) = Rez and Im(z +2kπi) = kπ +Imz, we have Loge z = z if
and only if k = 0 and this is the case if and only if −π < Imz ≤ π.
8. Show that the function f(z) = Log(−z) + iπ is a branch of logz in the domain D0 consisting of all
points in the plane except those on the nonnegative real axis. (In other words, show that e f(z) = z and
that f(z) is analytic on D0.)
Solution.
e f(z) = e Log(−z)+iπ = e Log(−z) e iπ = (−z)(−1) = z
Denote by D− the set of all complex numbers except 0 and those on the negative real axis. Log(ζ) is
analytic for all ζ ∈ D−. Hence f(z) = Log(−z) + iπ is analytic for all z obeying −z ∈ D−, which is
precisely all z obeying z ∈ D0.
9. Find a branch of log(z 2 +1) that is analytic at z = 0 and takes the value 2πi there.
Solution. Observe that z 2 +1 = (z +i)(z −i). Set f(z) = Log(z +i)+Log(z −i)+2πi. Then
e f(z) = e Log(z+i)+Log(z−i)+2πi = e Log(z+i) e Log(z−i) e 2πi = (z +i)(z −i)(1) = z 2 +1
Denote by D− the set of all complex numbers except 0 and those on the negative real axis. The domain
of analyticity of f(z) is the set of all z for which both z +i and z −i are in D−. This includes z = 0.
Finally,
f(0) = Log(0+i)+Log(0−i)+2πi = π π
2i− 2i+2πi = 2πi
10. For which values of τ do we have log τ(x) = lnx for all strictly positive real numbers x? Here lnx is the
real natural logarithm.
Solution. If x is a positive real number, then logx = lnx+2kπi with k ∈ Z. log τ x selects the value
of k for which τ < 2kπ ≤ τ +2π. This value of k is zero if and only if −2π ≤ τ < 0.
11. Let c be a nonzero constant. The multiple–valued function c z is given by c z = e zlogc . Show that by
selecting a particular value of logc we obtain a branch of c z that is entire. Find the derivative of such a
branch.
3

Solution. Pick any particular value of logc and call it log τ c. The function g(z) = log τ c z (i.e.
multiplication of z by the constant log τ c) is a polynomial and hence is entire. The function e zlog τ c is
the composition of e z , which is entire, with g(z) and hence is also entire.
d
dzezlog τ c = (logτ c) e zlogτ c
12. By definition, θ = tan −1 z is the solution of tanθ = z. Derive the formulae
(a) tan−1z = i i+z i
2 log i−z = 2
Solution. (a) For any z = ±i,
1−iz
d
log 1+iz , z = ±i (b) dz tan−1z = 1
1+z2, z = ±i
tanθ = z ⇐⇒ sinθ
2 e
cosθ = z ⇐⇒ 2i
iθ −e −iθ
⇐⇒ (1−iz)e iθ = (1+iz)e −iθ ⇐⇒ e −2iθ = 1−iz
1+iz
⇐⇒ θ = i 1−iz
2 log 1+iz
To get the form i i+z
2 log i−z
(b)
d i
dz 2
i+z i log i−z = 2
d
dz
e iθ +e −iθ = z ⇐⇒ e iθ −e −iθ = iz e iθ +e −iθ
⇐⇒ −2iθ = log 1−iz
1+iz
just multiply both the numerator and denominator of 1−iz
1+iz
i 1 −1 i
log(i+z)−log(i−z) = 2 i+z − i−z = 2
4
2i
(i+z)(i−z)
by i.
= − 1
−1−z 2 = 1
1+z 2