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Complex Variables . . . . . . . . . . . . . . . . . . . . . Exam One – Comments . . . . . . . . . . . . . . . . . . . . . Fall 2012
(1) Verify the following identity for the complex versions of sine and cosine:
sin(2z) = 2 cos z sin z.
Comments: We expand 2 cos z sin z as follows
iz
iz
e + e−iz
e − e−iz
2 cos z sin z = 2
2
2i
1 iz
= (e + e−iz ) (eiz − e−iz )
2i
1
= ((eiz )2 − (e−iz )2 )
2i
1
= (ei2z − e−i2z )
2i
= sin(2z).
1
dz where
−4
Γ
(a) Γ is the circle |z| = 1,
(b) Γ is the circle |z| = 5,
(c) Γ is the circle |z + 2| = 1, (d) Γ is the circle |z − 2| = 1.
Comments: We first expand 1/(z 2 − 4) in partial fractions:
(2) Evaluate the integral
Z
z2
z2
1
1
1
=A
+B
,
−4
z−2
z+2
1 = A(z + 2) + B(z − 2).
With z = 2, we find 1 = 4A or A = 1/4. With z = −2, we get 1 = −4B so B = −1/4. Note
A = −B.
(a) The integral is zero by Cauchy’s Theorem, since 1/(z 4 − 4) is analytic in the domain enclosed
by Γ.
(b)
Z
Z
1
1
1
dz =
A
+B
dz = 2Aπi + 2Bπi = 0
2
z−2
z+2
|z|=5
|z|=5 z − 4
since A = −B.
(c)
Z
|z+2|=1
(d)
Z
A
|z−2|=1
1
1
+B
z−2
z+2
A
1
1
+B
z−2
z+2
1
dz = 2Bπi = − πi.
2
dz = 2Aπi =
1
πi.
2
(3) Use the Cauchy-Riemann equations to verify that the function
f (z) = f (x + iy) = 2x2 − 2y 2 + 2xy + i(−x2 + y 2 + 4xy)
is analytic for all x and y . Find the value of the derivative f ′ (1 + i).
Comments: Let f (z) = u(z) + iv(z) = u(x, y) + iv(x, y), then the Cauchy-Riemann equations
are
ux = vy ,
uy = −vx ;
f ′ (z) = ux (z) + ivx (z).
1
In our case, u(x, y) = 2x2 − 2y 2 + 2xy and v(x, y) = −x2 + y 2 + 4xy. Then
ux = 4x + 2y,
uy = −4y + 2x,
vx = −2x + 4y,
vy = 2y + 4x.
We find that ux = vy and uy = −vx . Hence f (z) is analytic for all z. Further
f ′ (1 + i) = ux (1, 1) + ivx (1, 1) = 6 + 2i.
(4) (a) Find all four solutions to iz 4 + 8z = 0. It is acceptable to write the four distinct solutions in
polar form.
(b) Find all solutions to the equation cos z = 3/2. Write the solutions in the form x + iy .
Comments: (a) The equation iz 4 + 8z = 0 has the solution z = 0. To find the nonzero solutions,
cancel out the z to obtain iz 3 + 8 = 0. The solutions to this reduced equation are found by solving
z 3 = 8i; that is, by finding the cube roots of 8i. Write 8i in polar form 2eiπ/2 . Then its three cube
roots are
w0 = 2ei(π/2)/3 = 2eiπ/6 ,
w1 = 2ei(π/2+2π)/3 = 2ei5π/6 ,
w2 = 2ei(π/2+4π)/3 = 2ei3π/2 .
(b) Expand
cos z = 3/2
eiz + e−iz
= 3/2
2
eiz + e−iz = 3
(eiz )2 + 1 = 3eiz
(eiz )2 − 3eiz + 1 = 0.
The solutions to this quadratic are
√
√
3± 5
1
.
3± 9−4 =
e =
2
2
We note that both of the above solutions are positive real numbers. Next we get
"
#
√ !
√ !
√ !
3± 5
3± 5
3± 5
z = −i log
= −i ln
+ 2πik = −i ln
+ 2πk,
2
2
2
iz
k ∈ Z.
(5) (a) Use complex algebra to simplify the equation |z + 2| = |z + 1|. Use the simplified expression
to describe the set of solutions geometrically.
Comments: Rewrite |z + 2| = |z + 1| as |z + 2|2 = |z + 1|2 and expand:
|z + 2|2 = |z + 1|2
(z + 2)(z + 2) = (z + 1)(z + 1)
(z + 2)(z + 2) = (z + 1)(z + 1)
zz + 2z + 2z + 4 = zz + z + z + 1
z + z = −3
3
z+z
=−
2
2
3
ℜ(z) = − .
2
Hence the set of solutions is the vertical line given by ℜ(z) = −3/2.
2
5(b) Give a reason why the principal branch of the logarithm is NOT analytic on the portion of the
unit circle eit , where −π/4 ≤ t ≤ 5π/4. See the plot:
5(c) Give a branch of the logarithm that is analytic on the portion of the unit circle eit , where
−π/4 ≤ t ≤ 5π/4.
Comments: (b) The principal branch of the logarithm is not analytic on this arc because the principal
branch is NOT continuous along the negative real axis and this arc contains the point −1.
(c) Define the branch
log(z) = ln |z| + iθ,
−π/2 < t < 3π/2, z 6= 0.
This branch is analytic on the complement of the “negative imaginary axis;” that is, this branch is
analytic on the complement of the set {z ∈ C : ℜ(z) = 0 and ℑ(z) ≤ 0}.
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