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Aaron Morrisett
1.1
Monday, June 01, 2009 9:38 AM
1.2
Monday, June 01, 2009 12:44 PM
Aaron Morrisett Math 459 Homework # 2 1.3 Monday, June 01, 2009 8:37 PM
1.4
Monday, June 01, 2009 8:37 PM
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1.5
Wednesday, June 03, 2009 9:31 AM
2.1
Thursday, June 04, 2009 9:37 AM
2.2
Thursday, June 04, 2009 9:37 AM
2.3
Thursday, June 04, 2009 2:07 PM
2.6
Monday, June 08, 2009 3:43 PM
3.1
Tuesday, June 09, 2009 8:08 AM
3.2
Thursday, June 11, 2009 8:05 AM
4.3
Wednesday, June 17, 2009 9:47 AM
5.1
Thursday, June 18, 2009 9:31 AM
5.2
Friday, June 19, 2009 9:14 AM
5.3
Sunday, June 21, 2009 7:15 PM
cx
5.4
Monday, June 22, 2009 10:45 AM
5.5
Monday, June 22, 2009 2:07 PM
6.2
Friday, June 26, 2009 8:08 AM
6.3
Friday, June 26, 2009 11:42 AM
6.4
Monday, June 29, 2009 9:39 AM
6.5
Monday, June 29, 2009 11:55 AM
6.6
Tuesday, June 30, 2009 9:20 AM
Joseph Heavner Honors Complex Analysis Assignment 3 April 6, 2015
2.5 Reciprocal Function
1.) Find the image of the circle|z| =5 under the reciprocal function mapping.
This is classical inversion in a unit circle, in which the argument is invariant but the modulus is trans-formed to be the reciprocal of that of the domain. In other words, here our modulus becomes 1/5, making our image
The Circle|w| = 1
5
3.) Find the image of the semicircle|z| =3 , −π/4 ≤arg(z) ≤ 3π/4 under the reciprocal function
mapping.
Again, our function inverts the moduli to be 1/3, but here we too have reflection about the x-axis plays
a non-trivial role. In particular, this inverts (additive) the arguments to be−3π/4 ≤ arg(w) ≤ π/4.
Thus, we have The Semicircle|w| = 1 3 , − 3π 4 ≤arg(w) ≤ π 4
5.) Find the image of the annulus 1/3≤ |z| ≤2 under the reciprocal function mapping.
As with (1) reflection about the x-axis has no bearing on the image (fully circular regions are invariant under rotation). However, it remains that the moduli are inverted (multiplicative), thus we have
The Annulus1
2 ≤ |w| ≤3
7.) Find the image of the ray arg(z) =π/4 under the reciprocal function mapping.
Suppose we let z=reiθ, then our domain is given by r >0 , θ=π/4. (Note that r cannot be negative
in any case, and the case where r is zero includes all arguments and so is ignored here.) If we invert r
then we have 1/r>0, but because this includes the set of all possible r, we have an invariant (intuition
of the fact will here suffice, though a proof is accessible). Now, we also know that our mapping inverts
arguments, in particular θ=π/47−→θ0 = −π/4. In conclusion, we have
The Ray arg(w) = −π
4
9.) Find the image of the line y=4 under the reciprocal function mapping.
This problem fits a general form that states that a line y=k, under the reciprocal function, maps to the
circle|w+12k| = |1
2k|. With that in mind, the image is clearly
|w+1
8i| =
1 8 1
10.) Find the image of the line x= 16under the reciprocal function mapping.
Similar to (9) we have a problem of the form: find the image of the line x = k under the reciprocal
mapping, which has been shown to be answered by
|w− 1
2k| = |
1
2k|
Thus, we arrive at our answer
|w−3i| =3
11.) Find the image of the circle|z+1| =1 under the reciprocal mapping.
As mentioned in the Remarks on page 96 and can be seen from the results regarding have lines as
domains, this reciprocal mapping maps a circle of the form|z+2k1i| = |1
2k| to the line y = k. So, we
identify k=1/2, implying that we map to the line
y= 1
2
15.) Find the image of the set S under the mapping w=1/z onC∪ {∞}
. Figure 1: The Set S
We already know how to map lines to circles via the reciprocal function. So, we simply identify these
as the lines x= −2 and x= −1, which map to
|w+1 4| = 1 4 ; |w+ 1 2| = 1 2
We now check a point within the domain such as, say, z= −3/2, which maps to w= −2/3. Thus, we
have the set containing the point w= −2/3 and bounded by the circles
|w+1 4| = 1 4 ; |w+ 1 2| = 1 2 2
23.) Show that the image of the line x =k, x6= 0, under the reciprocal map defined on the extended complex plane is the circle
|w− 1
2k| = |
1
2k|
The vertical line x=k consists of all points z=k+iy such that x, y∈R where k6=0. So we may write
w= 1 k+iy = k k2+y2− y k2+y2i
Thus our real and imaginary parts of w= f(u, v)are
u= k
k2+y2 , v=
−y
k2+y2 , y∈R
But observe that v= −yuk , which implies
y= −vk
u (1)
And so, upon substituting into our initial expression for u, we have
u= k
k2+ (−vk
u )2
It is here advantageous to us to simplify and complete the square as follows
u= k k2+ (−vk u )2 k=u k2+ (−vk u ) 2 k=k2u+v 2k2 u uk=k2u2+v2k2 0=u2+v2−u k −1 2k 2 =u2−u k + − 1 2k 2 +v2 1 4k2 = u− 1 2k 2 +v2
Finally, by recognition, we have
|w−1
2k| = | 1 2k|
Note that we never made the restriction in (1) that u6= 0. This is because here we are working in the
extended complex plane.
2.6 Limits and Continuity
Note: In order to save time, the extreme pedantism and wordiness that is seen in the text has here been avoided. For instance, substitution to evaluate a limit may take one line rather than half a page.
1.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim
z→2i(z 2−z)
Let us simply substitute in z=2i as follows:
(2i)2−2i=4i2+2i= −4+2i
3.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim
z→1−i(|z| 2−i¯z)
Again, let us try substitution
|1−i|2−i(1−i) =2−i+i2=2+1−i= 3−i
5.) Use Theorem 2.6.1 and the properties of real limits on page 104 to evaluate lim
z→πi(e z)
Again, we need only substitute.
eiπ = −1
9.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim
z→2−i(z 2−z)
We try substitution as follows
(2−i)2−2+i=4−4i+i2+i−2=4−2−1−3i= 1−3i
11.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim
z→eiπ/4(z+ 1
z)
Yet again, we need only substitution for the following limit. In particular, lim
z→eiπ/4(z+ 1
z) =e
iπ/4+e−iπ/4=cos(π/4) +i sin(π/4) +cos(−π/4) +i sin(−π/4) = √2
13.) Use Theorem 2.6.2 and the basic limits (15) and (16) to compute lim
z→−i
z4−1
z+i
For the first time thus far, substitution will not suffice, for, obviously, it yields an indeterminate form
(note that L’Hopital’s rule applies only to the reals). However, if we simply recognize that z4−1 =
(z2−1)(z2+1) = (z−1)(z+1)(z+i)(z−i), then we have that
lim
z→−i
z4−1
z+i =z→−ilim(z+1)(z−1)(z−i) = (−i+1)(−i−1)(−2i) = 4i
19.) Consider the limit
lim
z→0
z
z
2
a.) What value foes the limit approach as z approaches 0 along the real axis? b.) What value does the limit approach as z approaches along the imaginary axis? c.) Do the answers from (a) and (b) imply that the limit exists? Explain.
d.) What value does the limit approach as z approaches along the line y=x?
e.) What can you say about the limit in general?
a.) Along the x axis we have z=x+0y=x, thus the limit is
lim x→0 x x 2 =lim x→01= 1
b.) Along the imaginary axis we have x=0, implying that z=iy, thus our limit becomes
lim y→0 yi −yi 2 =lim y→0(−1) 2= 1
c.) No. The limit must be the same along any of the infinitely (uncountably) many paths in the complex plane for it to exist in general. In other words, two, three, or even a trillion paths, while perhaps suggesting that the limit may exist and equal some value c, do not actually demonstrate that the limit is c. This must be shown using more general methods. However, if the limit along one path does not equal the limit along another, then we may say that the limit does not exist.
d.) If we approach along y=x then z=x+ix and
lim x→0 x+xi x−xi 2 = lim x→0 1+i 1−i 2 = 2i −2i = −1
e.) As discussed in part (c), the limit does not exist because the limit is dependent on path.
21.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute lim
z→∞
z2+iz−2
(1+2i)z2
Let us attempt to simplify our expression, as its current form yields an indeterminate form and there
is no natural idea of the top ”approaching infinity faster” than the bottom as there is inR (this is more
formally shown by dividing by the largest power of the variable in question), at least not when we have an imaginary number in question. So, upon simplification we have
(1−2i)z2+ (2+i)z− (2−4i) 5z2
Now that all factors are properly separated with the denominator a real expression of a complex vari-able, we may use our familiar laws of rational functions as apply in R. In particular, it is found that
lim z→∞f(z) = 1 5 − 2 5i
23.) Use (24) or (25), Theorem 2.2, and the basic limits (15) and (16) to compute lim
z→i
z2−1
z2+1
By substitution we have that, noting the nature of complex infinity (e.g. its lack of sign)
z2−1
z2+1 →
−2
0 → ∞
27.) Show that f is continuous at z0
f(z) =z2−iz+3−2i ; z0=2−i We have that f(2−i) =5−8i and similarly lim z→2−i(z 2−iz+3−2i) = (2−i)2−i(2−i) +3−2i=4−4i−1−2i−1+3−2i=5−8i
Therefore, by definition f is continuous at z0.
28.) Show that f is continuous at z0
f(z) =z3−1
z ; z0=3i
Similar to (27) we know that f(z0) = −80i3 and the limit at that point is the same by virtue of
substitu-tion, i.e. lim z→z0 f(z) = (3i)3− 1 3i =27i 3− 1 3i = −27i− 1 3i = −80 3i Thus f is continuous at z0. 6
29.) Show that f is continuous at z0
f(z) = z
3
z3+3z2+z; z0=i
Substitution here works for computing the limit, and substitution is equivalent in computation to
eval-uating f(z0), thus the two are equal. Here is the explicit calculation:
lim z→z0 f(z) = i 3 i3+3(i)2+i = −i −i−3+i = −i −3 = 1 3i And so f is continuous at z0.
31.) Show that f is continuous at z0
f(z) = ( z3−1 z−1 :|z| 6=1 3 :|z| =1 where z0=1
Clearly, f(1) =3 and so we simply try the limit as follows (note the expression we chose to evaluate –
this was done because it is equivalent to the value obtained along the other possible path) lim z→1 z3−1 z−1 =z→1lim (z−1)(z2+z+1) (z−1) =z→1lim(z 2+z+1) =3
Therefore, it has been demonstrated that f is continuous at z0=1.
35.) Show that f is discontinuous at z0
f(z) = z 2+1 z+i ; z0= −i Evaluating f(−i)yields (−i)2+1 −i+1 = ∅
Thus, the criteria for continuity are not met and f is discontinuous at z0.
37.) Show that f is discontinuous at z0
f(z) =Arg(z); z0= −1
The problem here is not that f does not exist but that in fact the limit does not exist, thus not meeting
the criteria for continuity, and so making f discontinuous at z0.
Let z be a point on the negative real axis, then Arg(z) =πbut we have that there are points{zn}such
that they are arbitrarily close to z but have their image under f has some nonzero imaginary part and
so the argument becomes arbitrarily close to−π. Thus, we have that the limit does not exist. (An e−δ
argument would make this more precise — Also, note that we must be careful with substitution in the case of poorly behaved functions like this)
39.) Show that f is discontinuous at z0 f(z) = ( z3−1 z−1 :|z| 6=1 3 :|z| =1 where z0=i
If we consider the approach along the imaginary axis then we have
lim z→i f(z) = i
3−i
i−1 =i
However, trivially we know that along the unit circle we have that the limit evaluates to 3. But, 36= i,
thus f is discontinuous at the point in question.
41.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is
continuous. z0
f(z) =Re(z)Im(z)
Let z=x+iy, then
f(z) =Re(x+iy)Im(x+iy) =xy
But, we know that the function f(z) = xy is continuous for arbitrary z, thus f is continuous on all of
C.
43.) Use Theorem 2.3 to determine the largest region in the complex plane on which the function f is continuous.
f(z) = z−1
z¯z−4
Let z=x+iy, then
f(z) = x+iy−1 (x+iy)(x−iy) −4 = x−1 x2+y2−4+i y x2+y2−4 Therefore, u(x, y) = x−1 x2+y2−4 ; v(x, y) = y x2+y2−4
These expressions are continuous on their domains. In particular, they are continuous for all(x, y) :
x2+y26=4, thus f is continuous for all z :|z| 6=2.
3.1 Differentiability and Analyticity
3.) Use definition 3.1 to find f0(z)where f(z) =iz3−7z2
The definition f0(z) = lim ∆z→0 f(z+∆z) − f(z) ∆z becomes f0(z) = lim ∆z→0 i(z+∆z)3−7(z+∆z)2−iz3+7z2 ∆z = lim ∆z→0 i(z3+3z2∆z+3z+∆z3) −7(z2+2z∆z+∆z2) −iz3+7z2 ∆z = lim ∆z→0 ∆z(3iz2+3iz∆z+i(∆z)2−14z−7∆z) ∆z = lim ∆z→03iz 2+3iz∆z+i(∆z)2−14z−7∆z =3iz2−14z Thus, f0(z) =3iz2−14z
5.) Use definition 3.1 to find f0(z)where f(z) =z−1z
f0(z) = lim ∆z→0 z+∆z− 1 z+∆z−z+1z ∆z = lim ∆z→0 z∆z+z2+1 z(∆z+z) = z 2+1 z2 =1+ 1 z2 Thus, f0(z) =1+ 1 z2 9
9.) Use alternate definition (12) to find f0(z)where f(z) =z4−z2 The definition mentioned is restated below for completeness:
f0(z0) = lim
z→z0
f(z) − f(z0)
z−z0
With that, we identify the pieces and substitute, then solve as follows on the next page: f0(z0) =z→zlim 0 z4−z2−z40+z20 z−z0 = lim z→z0 z4−z40 z−z0 +z 2 0−z2 z−z0 = lim z→z0 (z2+z20)(z−z0)(z+z0) z−z0 −(z+z0)(z0−z) z0−z = lim z→z0z 3 0+z20z+z0z2+z3−z−z0 =z30+z20(z0) +z0(z0)2+ (z0)3− (z0) −z0 =4z30−2z0
Note, however, that this is a general point, so we may replace z0, and so we arrive at our solution
f0(z) =4z3−2z
19.) The function f(z) = |z|2is continuous at the origin.
(a) Show that f is differentiable at the origin.
(b) Show that f is not differentiable at any point z6=0.
(a) If we take|z|to be zz then we can easily demonstrate this
f0(z) = lim ∆z→0 (z+∆z)(z+∆z) −zz ∆z = lim ∆z→0z ∆z ∆z+∆z+z
At this point we may stop, noting that if z=0, then all terms vanish and so the function is differentiable
with derivative equal to zero.
(b) Given the final limit of (a) we see that the term ∆z∆z is problematic. Noting that the expression can
be rewritten as an exponential. If this is done we note that the angle φ could be anything, regardless of
how arbitrarily close∆z is to zero. Thus, this portion of the limit justifies the lack of differentiability of
f(z)at any point z0∈C : z06=0.
21.) Show that f(z) =z is nowhere differentiable.
We consider the definition of the derivative, here letting∆z=w.
lim w→0 z+w−z w =w→0lim w w =x,y→0lim x−yi x+yi
This is clearly non-existent, for along the line x=0 it evaluates to−1, whereas it is 1 along y=0.
23.) Use L’Hopital’s rule to compute lim
z→i
z7+i
z14+1
If we rewrite the numerator as P(z)and the denominator as Q(z)then we see that f(z)must be analytic
at z=i, because it is the quotient of analytic functions (polynomials). Thus, we differentiate both P(z)
and Q(z). lim z→i 7z6 14z13 = 7(i)6 14(i)13 = −7 14i = i 2
27.) Determine the points at which f(z) = iz
2−2z
3z+1−i is not analytic.
We can take the derivative of the function to begin. To save paper, the work here will be somewhat ignored. Regardless, the conclusion is that
f0(z) = (−2+2i)(2+2i)z+3iz
2
((1−i) +3z)2
Now, we see if this is undefined anywhere, and indeed((1−i) +3z)2 = 0 at z = −13+ 13i, thus the
function cannot be analytic there. But, because the function’s derivative exists at all other points z∈C
we have the function is analytic everywhere except at z= −13+13i.
3.2 Cauchy–Riemann Equations
1.) Given that f is analytic, show that the Cauchy-Riemann equations are satisfied for f(z) =z3.
By expanding and grouping we rewrite f(z) = z3as f(z) = (x3−3xy2) +i(3x2y−y3) we take the
partial derivatives: ∂u ∂x =3x 2−3y2; ∂u ∂y = −6xy ; ∂v ∂x =6xy ; ∂v ∂y =3x 2−3y2
Thus the equations ∂u
∂x = ∂v ∂y and ∂u ∂y = − ∂v ∂x are satisfied.
3.) Show that f(z) =Re(z)is nowhere analytic.
If we let z=x+iy then f(z) =x, thus u(x, y) =x and v(x, y) =0. Therefore,
∂u ∂x =1 ; ∂u ∂y =0 ; ∂v ∂x =0 ; ∂v ∂y =0
Because the Cauchy-Riemann equations are not satisfied (in particular ∂u/∂x6=∂v/∂y) for no points in
the complex plane. So, we may conclude that f(z)is nowhere analytic.
7.) Show that f(z) =x2+y2is nowhere analytic.
u(x, y) =x2+y2and v(x, y) =0. Computing the partials of f :
∂u ∂x =2x ; ∂u ∂y =2y ; ∂v ∂x =0 ; ∂v ∂y =0
Therefore the Cauchy-Riemann equations are only satisfied at the point z = 0 in the complex plane.
However, there does not exist a domain R with z = 0 in R in which there all points in R satisfy the
Cauchy-Riemann equations. Therefore, f is nowhere analytic.
8.) Show that f(z) = x x2+y2+i y x2+y2 is nowhere analytic. u(x, y) = x x2+y2 and v(x, y) = y
x2+y2, which means that
∂u ∂x = y2−x2 (x2+y2)2 ; ∂u ∂y = −2xy (x2+y2)2 ; ∂v ∂x = −2yx (x2+y2)2 ; ∂v ∂y = x2−y2 (x2+y2)2
We observe that ∂u/∂x=∂v/∂y if and only if y= ±x where x6=0. However, ∂u/∂y= −∂v/∂x if and
only if x=0 where y 6=0 or y=0 where x6= 0. Clearly, these two conditions are mutually exclusive,
ensuring that the Cauchy-Riemann equations are not satisfied, implying that f is nowhere analytic.
9.) (a) Use Theorem 3.2.2 to show that f(z) = e−xcos y−ie−xsin y is analytic in an appropriate domain, and (b) find the derivative of f in said domain using (9) or (11).
(a) u(x, y) =e−xcos y and v(x, y) = −e−xsin y, therefore
∂u ∂x = −e −xcos y ; ∂u ∂y = −e −xsin y ; ∂v ∂x =e −xsin y ; ∂v ∂y = −e −xcos y
Clearly the Cauchy-Riemann equations are satisfied for all z ∈ C, making f analytic in all of the
complex plane.
(b) It is here advantageous to use ”Cartesian” (perhaps the term ”Argand” or rectangular is better here)
coordinates, so we compute f0(z)by using (9), which states that
f0(z) = ∂u ∂x +i ∂v ∂x = ∂v ∂y−i ∂u ∂y
Via substitution (note that here the book appears to have the incorrect answer) f0(z) = −e−xcos y+ie−xsin y
11.) (a) Use Theorem 3.2.2 to show that f(z) = ex2−y2cos(2xy) +iex2−y2sin(2xy) is analytic in an
appropriate domain, and (b) find the derivative of f in said domain using (9) or (11).
(a) u(x, y) =ex2−y2cos(2xy)and v(x, y) =2x2−y2sin(2xy), so
∂u ∂x =2e
x2−y2
(x cos(2xy) −y sin(2xy)); ∂u
∂y = −2e
x2−y2
(y cos(2xy) +x sin(2xy))
∂v ∂x =2e
x2−y2
(y cos(2xy) +x sin(2xy)); ∂v
∂y =2e
x2−y2
(x cos(2xy) −y sin(2xy))
It is now clear that f is an entire function, i.e. it is analytic throughout all of the complex plane.
(b) By formula (9) f0(z) = f(z)and in particular
f0(z) =excos(y) +iexsin(y)
17.) Find real constants a, b such that f(z) =3x−y+5+i(ax+by−3)is analytic.
u(x, y) =3x−y+5 and v(x, y) =ax+by−3, so ∂u ∂x =3 ; ∂u ∂y = −1 ; ∂v ∂x =a ; ∂v ∂y =b
It is therefore clear that
a=1, b=3
19.) Show that f(z) =x2+y2+2ixy is not analytic at any point but is differentiable along the x-axis,
and (b) use (9) or (11) to compute the derivative along the axis.
(a) u(x, y) =x2+y2and v(x, y) =2xy, thus
∂u ∂x =2x ; ∂u ∂y =2y ; ∂v ∂x =2y ; ∂v ∂y =2x
The first Cauchy-Riemann equation is universally satisfied, whereas the second is satisfied if and only
if y=0. Thus, while f is differentiable on the x-axis, it is not analytic, for there exists no neighborhood
in which the Cauchy-Riemann equations are satisfied.
(b) Via (9), f0(z) =2x+i2y, which, along the x-axis where y=0 means that
f0(z) =2x
21.) (a) Show that f(z) =x3+3xy2−x+i(y3+3x2y−y)is not analytic at any point but is
differen-tiable along the coordinate axes, and (b) use (9) or (11) to compute the derivative along the axes.
(a) u(x, y) =x3+3xy2−x and v(x, y) =y3+3x2y−y, meaning that
∂u ∂x =3x 2+3y2−1 ; ∂u ∂y =6xy ; ∂v ∂x =6xy ; ∂v ∂y =3y 2+3x2−1
The first Cauchy-Riemann equation is satisfied for all z ∈ C, whereas the second is satisfied only if
x =0, y=0, or x=y=0. Similar to (19), there exists no neighborhood in which the Cauchy-Riemann
equations are satisfied, therefore f is not analytic, but the Cauchy-Riemann equations are satisfied on the coordinate axes, making them differentiable there.
(b) In particular,
f0(z) =3x2−1 along the x−axis f0(z) =3y2−1 along the y−axis
Joseph Heavner Honors Complex Analysis Assignment 1 December 26, 2014
1.1 Complex Numbers and Their Properties
2.) Write the given number in the form a+bi.
(a) 2i3−3i2+5i (b) 3i5−i4+7i3−10i2−9 (c) 5 i + 2 i3 − 20 i18 (d) 2i 6+ 2 −i 3 +5i−5−12i
(a) Given the identity i2 = −1 we have the following re-expression 2(−i) −3(−1) +5i, which simplifies
further to 3+3i
(b) Because the powers of i are periodic with a cycle of 4 we can determine the value of in for any n∈Z+
by simply taking in mod 4. Using this, we have the following simplification 3(i) − (1) +7(−i) −10(−1) −9
which leads us to our solution 0−4i
(c) Let us rewrite this as 5
i + 2 (−i)− 20 −1 = 5i i2 + 2i −i·i+20= −5i+2i+20= 20−3i
(d) Again, we begin by rewriting our exponents, but this time we too distribute exponentiation over a
quotient as follows 2(−1) +(−i)233+5(1i) −12i= −2+8ii2 +5ii2 −12i= −2−8i−5i−12i= −2−25i
23.) Use the binomial theorem to write(−2+2i)5in a+bi form.
Recall the binomial theorem.
Binomial Theorem:Let x, y∈C, then
∀n∈Z+ :(x+y)n = n
∑
k=0 n k xn−kykIn our situation, this is
(−2+2i)5= 5
∑
k=0 5 k (−2)5−k (2i)k Expanding this out yields the following
1· (−2)5(2i)0+5· (−2)4(2i)1+10· (−2)3(2i)2+10· (−2)2(2i)3+5· (−2)1(2i)4+1· (−2)0(2i)5
Upon simplication (term by term, keeping symmetry in mind) we have−32+160i−320i2+320i3−160i4+
32i5, which, using the definition of i gives−32+160i+320−320i−160+32i, which yields our answer
128−128i
35.) Show that z1= − √ 2 2 + √ 2
2 i satisfies the equation z
2+i=0. Find an additional solution, z
2.
To verify the suggested solution is indeed valid we need only substitute and finish with an identity, thus − √ 2 2 + √ 2 2 i !2 +i=0 − √ 2 2 !2 +2 − √ 2 2 · √ 2 2 i ! + √ 2 2 i !2 = −i 1 2+2(−.5i) + i2 2 = −i 1 2 − 1 2−i= −i −i= −i
Now, to find an additional solution we use a clever trick. Because the squaring function transforms all negatives into positives and because of the nature of i, we know that the solution set here will be
x= {a+bi ,−a−bi}
So, we must show, like before, that we can satisfy our equation, but this time with z2=
√ 2 2 − √ 2 2 i √ 2 2 − √ 2 2 i !2 +i=0 √ 2 2 !2 +2 − √ 2 2 · √ 2 2 i ! + − √ 2 2 i !2 = −i 1 2+2(−.5i) + i2 2 = −i 1 2 − 1 2−i= −i −i= −i
Given that, we have our answer via recognition and, finally, direct verification:
z2= √ 2 2 − √ 2 2 i
46.) Think of an alternative solution to Problem 24. Then without doing any significant work, evaluate (1+i)5404.
For problem 24 an elegant solution would be as follows:(i+1)2=1+2i−1=2i so
(i+1)8=(i+1)24= (2i)4=24·i4=16 Similarly, for the more daunting problem at hand we have
(1+i)5404=(1+i)22702= (2i)2702= (i2)1351·22702= −22702
49.) Assume for the moment that√1+i makes sense in the complex number system. Demonstrate the validity of the equality
√ 1+i= r 1 2+ 1 2 √ 2+i r −1 2+ 1 2 √ 2
We may not have reached roots in our text yet, but we can still use exponents and eliminate the root, leaving us with something that our text has discussed. In particular, here is the verification, ignoring the negative root for now
√ 1+i2= r 1 2 + 1 2 √ 2+i r −1 2+ 1 2 √ 2 !2 i+1= r 1 2 + 1 2 √ 2 !2 +2 i r −1 2+ 1 2 √ 2 ! r 1 2 + 1 2 √ 2 !! + i r −1 2+ 1 2 √ 2 !2 i+1= 1 2+ 1 √ 2 +2 1 2i +1 2 − 1 √ 2 1+i=1+i
Because we have reached an identity and our steps are reversible, we have successfully shown that the equality holds. Typically, one would ”clean up” this proof by starting from first principles and working backwards, but this is more transparent with respect to our methodology.
1.2 Complex Plane
3.) Graph z1=5+4i, z2= −3i, 3z1+5z2, and z1−2z2 as vectors.
Here I will manually draw these rather than use software, though the latter choice would look nicer. First,
however, we will find explicit expressions for the third and fourth graphs. In particular 3(5+4i) +5(−3i) =
15−3i and(5+4i) −2(−3i) =5+10i.
6.)
(a) Plot the points z1= −2−8i, z2=3i, and z3= −6−5i.
(b) The points in (a) determine a triangle with vertices z1, z2, z3. Express each side of the triangle as a
difference of vectors.
a.) Again, we choose to manually plot the points
b.) The side connecting points z1and z2is z2−z3, the side connecting z1and z3 is z3−z1, and the side
connecting z2and z3is z2−z1. This can be best seen geometrically and can be expressed as follows
−−→
z1z2=z2−z1 , −−→z1z3=z3−z1 , −−→z2z3=z3−z2
7.) In problem 6, determine whether the points z1, z2, z3are the vertices of a right triangle.
In order for a triangle to be a right triangle it must obey the Pythagorean theorem a2+b2=c2where a, b, c
are the lengths of the three sides of the triangle. So, we must determine the distance from z1to z2, from z1
to z3, and from z2to z3. This goes as follows
d1= |z2−z1| = q (0− (−2))2+ (3− (−8))2=√125 and d2= |z3−z1| = q (−6− (−2))2+ (−5− (−8))2=√25=5 and d3= |z3−z2| = q (−6− (0))2+ (−5−3)2=√100=10
Now, finally we have that
52+102=√1252
So, we have verified that the vertices form a right triangle.
11.) Find the modulus of
z= 2i
3−4i
We have to convert this to a+bi form as follows
2i 3−4i· 3+4i 3+4i = − 8 25+ 6i 25 Thus, we can take
|z| =q(−8/25)2+ (6/25)2=q(64/625) + (36/625) =√4/25= 2
5
17.) Describe the set of points z in the complex plane that satisfy
Re((1+i)z−1) =0
For z=a+bi we have
Re((1+i)(a+bi) −1) =Re(a+bi+ai−b−1) =a+b−1
Thus, we have the line a+b−1 = 0 or, equivalently, x−y = 1, which can be rewritten in the familiar
notation y=x−1.
19.) Describe the set of points z in the complex plane that satisfy |z−i| = |z−1|
Rewriting this by definition for z=x+iy we get
q
x2+ (y−1)2=q(x−1)2+y2
x2+y2−2y+1=x2−2x+1+y2
−2y= −2x
y=x
Thus, this set of points describes the line in the complex plane
y=x
23.) Describe the set of points z in the complex plane that satisfy |z−1| =1
This can be explained a few ways. One would be to set this up in terms of trigonometric functions and make the connection between the parametric equations governing the unit circle. Another would be to geometrically think of this in terms of the locus of all points 1 away from the center, which is at 1. More analytically, however, we can go through definitions until we get to the standard definition of the circle in
the Cartesian plane with radius r and center(h, k)
(x−h)2+ (y−k)2=r2
So, we use our definitions and algebra as follows to prove this is indeed what it is. For z=x+iy we have
q
(x−1)2+y2=1
(x−1)2+y2=1
This describes a circle with radius r=1 and center(1, 0).
27.) Establish the simultaneous inequality
If|z| =2, then 8≤ |z+6+8i| ≤12
Let z=x+iy, and we note that|z| =2 =⇒ x2+y2=4. Now, we split our modulus in two, with|z| =2,
thus
8≤2+ |6+8i| ≤12 =⇒ 8≤12≤12
By the triangle inequality, this must be the maximum (the least upper-bound AKA limit superior) of the
inequality. Similarly, we have that|z1| − |z2| ≤ |z1+z2|, thus the following expression yields the minimum
(greatest lower bound or limit inferior)
8≤ |6+8i| −2≤12 =⇒ 8≤8≤12
It has been suggested that perhaps it would be better form to combine these inequalities into one expression.
37.) Under what circumstance does|z1+z2| = |z1| + |z2|?
Let z1=a+bi and z2=x+iy then we have to find under which conditions
q
(a+x)2+ (b+y)2=pa2+b2+qx2+y2
So, we first square both sides as follows
(a+x)2+ (b+y)2=2pa2+b2qx2+y2+a2+b2+x2+y2
a2+2ax+x2+b2+2yb+y2=2pa2+b2qx2+y2+a2+b2+x2+y2
2(ax+yb) =2pa2+b2qx2+y2
ax+yb=pa2+b2qx2+y2
Re(z1)Re(z2) +Im(z1)Im(z2) = |z1||z2| = |z1z2|
Thus,|z1+z2| = |z1| + |z2|if and only if Re(z1)Re(z2) +Im(z1)Im(z2) = |z1z2|
Remark: A concrete example of this would be if|z1| =0 and/or|z2| =0. Challenge: Think of this
geometri-cally with vectors.
38.) Using the complex variable z, find an equation of a circle in the complex plane where θ is measured in radians from the positive x-axis.
The following equation defines a circle of radius r centered at some point z0in the complex plane
|z−z0| =r
This comes directly from the definition of a circle being the set of all points some distance d away from some center point. More precisely, as in (23), we can go through definitions to get the familiar formula for a circle, thus proving our proposition
q
(x−a)2+ (y−b)2=r
(x−a)2+ (y−b)2=r2
Which is our normal formula for a circle centered at(a, b)with radius r. Note that we define our complex
variable as z=x+iy and our complex number as z0=a+bi.
39.) Describe the set of points z in the complex plane that satisfy z=cos θ+i sin θ, where θ is measured
in radians from the positive x-axis.
By Euler’s identity we have that eiθ = cos θ+i sin θ, so we could say that it would be the set of all points
that can be produced from that. However, more specifically, we have that the this set of points can take on any complex number with modulus one by virtue of our representation of complex numbers as polar forms (exponential or trigonometric/circular). In other words, the set of all points satisfying this equality forms the unit circle in the complex plane.
This same figure and variants of it have already been featured multiple times in this assignment, but here we provide a quick algebraic verification of the proposition above. Because this is a parameterization such
that x = cos θ and y= sin θ we have that the set z of all numbers for x, y ∈ R such that z= {x+iy}. In
other words, any complex number, with the exception, which we hadn’t before mentioned, that x, y≤1 (a
fact obtained a priori by parametric equations and the images of sine and cosine) is represented by this. 7
49.) In this problem we will start you out in the proof of the first property|z1z2| = |z1||z2|in (3). By the
first result in (2) we can write|z1z2|2 = (z1z2) (z1z2). Now use the first property in (2) of section 1.1 to
continue the proof.
The property aforementioned is that
z1z2=z¯1z¯2
So, we proceed as follows
|z1z2|2= (z1z2) (z1z2)
= (z1z2) (z¯1z¯2)
= |z1|2|z2|2
Where the last step is obtained by the fact that|z|2 = z¯z. Now, we may root both sides of our equation to
obtain the desired identity, noting that|z1|,|z2| ∈R+.
q
|z1z2|2=
q
|z1|2|z2|2
|z1z2| = |z1| |z2|
50.) In this problem we guide you through an analytical proof of the triangle inequality (6).
Since|z1+z2| and |z1| + |z2| are positive real numbers, we have |z1+z2| ≤ |z1| + |z2| if and only if
|z1+z2|2≤ (|z1| + |z2|)2.
(a) Explain why|z1+z2|2= |z1|2+2Re(z1z2) + |z2|2.
(b) Explain why(|z1| + |z2|)2= |z1|2+2|z1z2| + |z2|2
(c) Use parts (a) and (b) along with the results in Problem 46 to derive (6).
a.) This derivation is pretty straightforward for z1=x+iy and z2=a+bi. Below is the work.
(x+a)2+ (y+b)2=x2+2ax+a2+y2+2by+b2
=a2+b2+2(ax+by) +x2+y2
= |z1|2+2Re(z1z2) + |z2|2
b.) Similarly, we get this result by mere algebraic manipulation without much thought. q
x2+y2+pa2+b2
2
=2pa2+b2qx2+y2+a2+b2+x2+y2= |z
1|2+2|z1z2| + |z2|2
c.) Given those results and the fact that|Im(z)| ≤ |z|and|Re(z)| ≤ |z|we can prove the triangle inequality,
which geometrically states that the length of two sides of a triangle must be greater than the remaining side.
Taking (a) and (b) we see that, because 2Re(z1z2) ≤2z1z2by the result in Problem 46,
|z1+z2|2≤ (|z1| + |z2|)2
Thus, it easily follows that, upon square-rooting both sides, |z1+z2| ≤ |z1| + |z2|
1.3 Polar Form of Complex Numbers
1.) Write z=2 in polar form, first using an argument θ6=Arg(z)and then using θ=Arg(z).
First, we find the modulus|2| =p
(2)2+ (0)2=2 and the principle argument θ=arctan(0/2) =0. Now,
using our polar form we have
2cis(2π)
and 2cis(0)
Note our introduction of the ”cis” notation for polar form. The convenience of this notation makes it op-timal, especially while we have not yet been introduced formally to the exponential form of a complex number. However, notations will be varied throughout for no reason other than diversity.
3.) Write z= −3i in polar form, first using an argument θ6=Arg(z)and then using θ=Arg(z).
Our modulus isp(0)2+ (3)2 = 3 and our argument is arctan −3
0
= −π
2 +2πk , k ∈ Z. Thus, the
following two are polar forms of z, with the first using an arbitrary argument (co-terminal to the principle argument) and the second being the principle argument.
z= ( 3cos −π 2 +i sin −π 2 3cos 3π 2 +i sin 3π2
7.) Write z= −√3+i in polar form, first using an argument θ 6=Arg(z)and then using θ=Arg(z).
Our modulus is|z| = q(−√3)2+ (1)2 = 2 and our argument is arctan(−1/√3) = 5π/6+2πk, so we
have our two answers
z=
(
2 cis(−7π/6)
2 cis(5π/6)
9.) Write z= −1+i3 in polar form, first using an argument θ6=Arg(z)and then using θ =Arg(z).
First we need z in a+bi form, so we begin by multiplying by the conjugate on both top and bottom then
we simplify −3−3i 2 = − 3 2− 3i 2
Now we find the modulus |z| = p
(−3/2)2+ (−3/2)2 = 3√2
2 and principle argument tan−1
−3 2 −3 2 =
arctan(1) = −3π4. Thus, we have our two answers
z= (3√2 2 cos −3π4 +i sin −3π 4 3√2 2 cos 5π4 +i sin 5π4 9
11.) Use a calculator to write z= −√2+√7i in polar form, first using an argument θ6=Arg(z)and then
using θ=Arg(z).
As before we find the modulus to be q
(√7)2+ (−√2)2=3 and, with the aid of a calculator, we find our
argument to be arctan(−√7/√2) ≈2.06168, making our solution
3 cis(8.34486)
and
3 cis(2.06168)
13.) Write the complex number z whose coordinates(r, θ)are(4,−5π/3)in the form a+bi.
In other words, z = 4 cis(−5π/3), which is the same as 4 cos(−5π/3) +4i sin(−5π/3) = 4(1/2) +
4i(√3/2). So, our final answer is
2+2√3i
15.) Write the complex number
z=5 cos7π 6 +i sin 7π 6
in the form a+bi.
Our trigonometric form is already partitioned into two parts, a and b, upon distribution, thus we have
a=5 cos7π6 = −5
√ 3
2 and b=5 sin7π6 = −52, so we have
z= −5 √ 3 2 − 5 2i
19.) Use (6) and (7) to find z1z2and z1/z2in the form a+bi where
z1=2 cosπ 8 +i sin π 8 , z2=4 cos3π 8 +i sin 3π 8
This requires few words, so here are the computation via (6) and (7):
z1z2=8 cis(π/2) =8 cos(π/2) +8i sin(π/2) = 8i
and z1
z2
= 1
2cis(−π/4) =.5 cos(−π/4) +.5i sin(−π/4) =
1 2 √ 2 2 + 1 2i √ 2 2 = √ 2 4 − √ 2 4 i 10
21.) Write z in polar form. Then use either (6) or (7) to obtain the polar form of the given number. Finally,
write the polar form in the form a+bi.
z= (3−3i)5+5√3i
I feel as though this question may be incorrectly written, or at least obscurely phrased. Anyway, based on the answer key in the back of the text, the question is fairly forward with the exception that we split the
complex number z into its divisors z1and z2.
For z1we have modulus p32+ (−3)2 =
√
18 and argument arctan(−1) = 7π/4. For z2we have
mod-ulus q
52+ (5√3)2 = 10 and argument arctan(√3) = π/3. So, we have z
1 =
√
18 cis(7π/4) and
z2=10 cis(π/3), meaning that our whole number
z=z1z2=10
√
18 cis(π/12)
In order to go back to a+bi form we simply distribute r and evaluate our trigonometric functions
10√18 cos(π/12) +10
√
18i sin(π/12) = 15(
√
3+1) +15i(√3−1)
23.) Write z in polar form. Then use either (6) or (7) to obtain the polar form of the given number. Finally,
write the polar form in the form a+bi.
z= −i
1+i
Similar to before, we let z= z1/z2, where z1= −i and z2=1+i. For z1the modulus isp02+ (−1)2 =1
and argument is (from pure geometry) 3π/2, and for z2the modulus is
√
12+12= √2 and the argument
is arctan(1) =π/4. Thus, z= √cis(3π/2) 2 cis(π/4) = √ 2 2 cis(5π/4)
To get to a+bi form we do as follows
√ 2 2 cos(5π/4) + √ 2 2 i sin(5π/4) = − 1 2− i 2 25.) Use (9) to compute 1+√3i9
If we take the modulus and argument of our z = 1+√3i to convert it into trigonometric form, then we
can easily determine the ninth power of the number via de Moivre’s formula. The modulus is given by q
(1)2+ (√3)2 = 2 and the argument arctan(√3) = π/3, so, in polar form, z = 2 cis(π/3). Given z,
finding z9is simply applying a formula as follows
z9= (2 cis(π/3))9=29cis(3π) =512(−1) = −512
31.) Write z in polar form and then in a+bi form. z=cosπ 9 +i sin π 9 12h 2cosπ 6 +sin π 6 i5
Using de Moivre’s formula we get
z= (cos(12π/9) +i sin(12π/9)) [32(cos(5π/6) +i sin(5π/6))]
Now, multiplying our polar forms we have
z=32 cis(13π/6)
Then, distributing r and evaluating our sine and cosine functions, we have
z=32 cos(13π/6) +32i sin(13π/6) =32(√3/2) +32i(1/2) = 16√3+16i
33.) Use de Moivre’s formula (10) with n=2 to find the trigonometric identities for cos 2θ and sin 2θ.
This classic exercise is fairly straightforward, we simply take de Moivre’s formula for n=2 as follows
(cos(x) +i sin(x))2=cos(2x) +i sin(2x)
Now we expand the binomial on the left hand side to get
cos2(x) −sin2(x) +2i sin(2x)cos(2x) =cos(2x) +i sin(2x)
Now, we simply take Re(z)on each side to get our expression for cosine and we take Im(z)on both sides
to get the expression for sine. In particular, we find that
cos(2x) =cos2(x) −sin2(x)
and
sin(2x) =2 sin(x)cos(x)
35.) Find the positive integer n for which the following holds √ 3 2 + 1 2i !n = −1
Because Chapter 1.4 is about roots, we unfortunately cannot simply take the roots of−1 and check. Instead,
we expand out binomials, first by brute force, then by the binomial theorem. This is still fairly tedious, but
it will do, especially because the n needed here is quite small. For n =1 we have just the result inside the
parentheses, for n = 2 we double-distribute, yielding i
√ 3
2 . For greater n we choose to use the binomial
theorem instead. For n=3 we have
3
∑
k=0 3 k √ 3 2 !3−k 1 2i k =iSimilarly, for n=4 and n=5 we have, respectively,
4
∑
k=0 4 k √ 3 2 !4−k 1 2i k = −1 2+ i√3 2 and 5∑
k=0 5 k √ 3 2 !5−k 1 2i k = − √ 3 2 + i 2And finally, we have that for n=6 the equality is satisfied, because
6
∑
k=0 6 k √ 3 2 !6−k 1 2i k = −1Thus, for n=6 the equality holds. (This could have been done alternatively with de Moivre’s theorem.)
39.) Suppose that z1=r(cos θ+i sin θ). Describe geometrically the effect of multiplying z1by a complex
number of the form z2=cos α+i sin α, when α>0 and when α<0.
Because the second complex number has modulus one, multiplying the moduli simply yields r, so the
vector used to describe z1z2is equal in length to that of z1(i.e. no scaling occurs). On the other hand, in
general, we have that θ 6= α, which means that our angle changes (indeed, θ0 = θ+α, where the prime
notation is reserved, not for differentiation, but rather for the argument of z1z2). For any alpha we clearly
have a rotation due to the addition of arguments. But, for positive alpha we have a larger angle measured
from the positive x =Re(z)axis; in other words, for positive alpha we have a counterclockwise rotation
of z1by α. Similarly, for negative alpha we have a clockwise rotation of z1by a factor of α.
42.) Are there any special cases in which Arg(z1z2) =Arg(z1)Arg(z2)? Prove your assertions.
We know that Arg(z1z2) = Arg(z1) +Arg(z2), therefore the only way we can have this is if Arg(z1) +
Arg(z2) =Arg(z1)Arg(z2). For z1 = x+iy and z2 = a+bi we have(y/x) + (b/a) = (y/x)(b/a). If we
let X = y/x and A = b/a then we have A+X = AX, which has solutions A, X = 2 , A, X = 0. So, in
other words, if y, b =0, i.e. the numbers are real, then the equality holds. Similarly, if y/x , b/a=2, then
the equality holds. The first case seems to have some important significance, the second seems less notable. Challenge: Think in terms of angle magnitudes and try to generalize.
1.4 Powers and Roots
1.) Use (4) to compute all roots of z. Give the principal nth root in each case. Sketch the roots
w0, w1,· · ·, wn−1on an appropriate circle centered at the origin.
z= (8)1/3
In this case the principal nth root is simple (the one learned in all elementary algebra classes) and evaluates
to 2 by the fact that 23 = 8. For the other roots we need more powerful techniques found in our text. In
particular, we begin by finding 8 in polar form. The modulus is 8 and argument 2π, from pure geometry.
Therefore, given our formula for roots for n=3, we find
(8)1/3=2 cis(0), 2 cis(2π/3), 2 cis(4π/3) = 2 , −1±√3i
9.) Use (4) to compute all roots of z. Give the principal nth root in each case. Sketch the roots
w0, w1,· · ·, wn−1on an appropriate circle centered at the origin.
−1+√3i1/2
First, we write z in polar form. The modulus is q
(−1)2+ (√3)2=2 and argument arctan(−√3) =2π/3.
So, from our root formula for n=2 we have the following (with the first being the principal root):
√ z=√2 cis(π/3), √ 2 cis(−2π/3) = √ 2 2 + √ 6 2i , − √ 2 2 − √ 6 2i ; ≈.7071+1.2247i , −.7071−1.2247i 14
15.)
(a) Verify that(4+3i)2=7+24i.
(b) Use part (a) to find the two values of(7+24i)1/2.
a.) We expand the binomial as follows, noting the definition of the imaginary unit
(4+3i)2=16+24i+9i2=16−9+24i=7+24i
b.) By virtue of the square-root function being inverse to the squaring function, with the caveat that there
are two square-roots of a number (not the principle root), one positive and one negative. Thus,(7+24i)1/2
is not only 4+3i but also−(4+3i), making our simplified solution set
{4+3i , −4−3i}
17.) Find all solutions of the equation
z4+1=0
By algebraic manipulation we have that the problem is to find the fourth roots of -1, which is fairly trivial when given our formula for roots
n √ z= √nr cos θ+2kπ n +i sin θ+2kπ n
Given that, we must only find the modulus and argument of z= −1 to proceed. Arg(−1) =tan−1 0−1 =π
and| −1| =p (−1)2+ (0)2=1, thus 4 √ −1=√41 cos π+2kπ 4 +i sin π+2kπ 4
which yields cos(π/4) +i sin(π/4) , cos(3π/4) +i sin(3π/4) , cos(5π/4) +i sin(5π/4) , cos(7π/4) +
i sin(7π/4)and simplifies to our final answer
4 √ −1= ± √ 2 2 ±i √ 2 2 15
19.)
(a) Show that the n nth roots of unity are given by
(1)1/n=cos2πk
n +i sin
2πk
n , k=0, 1, 2,· · ·, n−1
(b) Find the n nth roots of unity for n=3, 4, 5.
(c) Carefully plot the roots of unity found in (b). Sketch the regular polygons formed with the roots as vertices. [Hint: See (ii) in the Remarks.]
a.) This is trivial by our roots formula; we must only find the argument and modulus of z =1 as follows
Arg(1) =01 =0 and|1| =p
(0)2+ (1)2=1. Thus, plugging in θ =0 and r=1 into our formula we get
n √ 1=cos2πk n +i sin 2πk n for k=0, 1,· · ·, n−1.
b.) For n=3 we have, by substitution:
cis(0) , cis 2π 3 , cis 4π 3 = 1 , −1 2+ √ 3 2 ,− 1 2− √ 3 2
Similarly, for n=4 we get:
cis(0), cis(π/2), cis(π), cis(3π/2) = 1 , i ,−1 ,−i
And, finally for n=5 we have:
cis(0), cis(2π/5), cis(4π/5), cis(6/pi/5), cis(8π/5)
=
1 , 0.30902+0.95106i ,−0.80902+0.58779i ,−0.80902−0.58779i , 0.30902−0.95106i
c.)
22.) Consider the equation (z+2)n+zn = 0, where n is a positive integer. By any means, solve the
equation for z when n=1. When n=2.
For n=1 this simplifies to the linear equation
2z+2=0 =⇒ z= −1 For n=2 we have (z+2)2+z2=0 z2+4z+4+z2=0 2z2+4z+4=0 2(z2+2z+2) =0
Now, by the quadratic formula, which, derived from completing the square, states that for any second
degree polynomial ax2+bx+c=0 the following holds true
x= −b±
√
b2−4ac
2a we have the following
z= −2±p2 2−4(1)(2) 2(1) = −2±i√4 2 = −1±i 17
23.) Consider the equation in (22).
(a) In the complex plane, determine the location of all solutions z when n=5
[Hint: Write the equation in the form[(z+2)/(−z)]5=1 and use part (a) of Problem 19.]
(b) Reexamine the solutions of the equation in Problem 22 for n=1, 2. Form a conjecture as to the location
of all solutions of(z+2)n+zn =0.
a.) As suggested, we rewrite the equation as
z+2
−z
5
=1 =⇒ 11/5= z+2
−z
We know from (19) that the fifth roots of unity are
(1)1/5=cis(0), cis(2π/5), cis(4π/5), cis(6π/5), cis(8π/5)
So, we have five equations to solve, most of which involve trigonometric functions whose outputs are, well, messy. They are as follows.
z+2 −z =1 =⇒ z= −1 & z+2 −z ≈.30902+.95106i =⇒ z= −1+ q 5−2√5≈ −1+0.726541i & z+2 −z ≈.30902−.95106i =⇒ z= −1− q 5−2√5≈ −1−0.726541i & z+2 −z ≈ −.80902−.58779i =⇒ z= −1− q 5+2√5≈ −1−3.07767i & z+2 −z ≈ −.80902+.58779i =⇒ z= −1+ q 5+2√5≈ −1+3.07767i
b.) Thus far all solutions have been on the line Re(z) =x= −1, i.e. all roots ω0, ω1,· · ·, ωnhave real part
−1. So, we conjecture the following:
Conjecture: All solutions to(z+2)n+zn =0 lie on the line x= −1, where Re(z) =x.
27.) The vector given in Figure 1.4.3 represents one value of z1/n. Using only the figure and trigonometry
— that is, do not use formula (4) – find the remaining values of z1/n when n= 3. Repeat for n= 4 and
n=5.
The figure shows one root to be
ω0= −2
Thus all of our other roots will be the same distance from the origin (i.e. lie on the same circle), but rotated
by 2π/3. Thus, ω1is the number with modulus 2 and argument π/3, that is
ω1=2cis(π/3) =1+
√ 3i
Furthermore, ω2has modulus 2 and argument 5π/3, thus
ω2=2cis(5π/3) =1−
√ 3i
We know that, if−2 is a third root of z, we can obtain z by cubing−2. In other words, z= −8. We still avoid
using any formulas, but we note that the modulus of this number is 8, therefore the fourth root will possess
modulus 23/4. Furthermore, the argument is θ/n where θ is the argument of z and n is the index of the root,
thus for n = 4 we have argument π/4. So, we have the following roots, all with the same argument but
separated by a factor of π/2:
23/4cis(π/4), 23/4cis(−π/4), 23/4cis(3π/4), 23/4cis(−3π/4)
This would suffice for a drawing, but we will also convert to a+bi form, yielding
± √ 2 2 ± √ 2 2
Similarly, for n=5 we have modulus 23/5and argument π/5. So, our roots are, all separated by an angle
of 2π/5, as follows
23/5cis(π/5), 23/5cis(3π/5), 23/5cis(π), 23/5cis(−π/5), 23/5cis(−3π/5)
In a+bi form that is approximately
1.2262±0.8909i ,−0.4684±1.4415i ,−1.5157
Note: This may be flawed; I will have to look at the work and interpretation again later.
28.) Suppose n denotes a nonnegative integer. Determine the values of n such that zn =1 possesses only real solutions. Defend your answer with sound mathematics.
This question regarding roots of unity is deceivingly important and not entirely straightforward. Solving
our equation we get that z=11/n, so we use our nth root formula for 1, which has modulus 1 and argument
0, thus we have that the nth roots are n √ 1=cos 2πk n +i sin 2πk n
So, for answers in the reals we look at the image of the trigonometric functions sine and cosine, which,
for both, is[−1, 1]. Thus, for a purely real answer we needn’t worry about the cosine (this will always be
real) and instead turn our attention to sine, which is multiplied by i. Now, we must find which values of p
satisfy the requirement(i·p) ∈ R. Now, we know that p=ni , n∈ R satisfy this by definition of i and it
also follows that p=0 works because 0i=0∈R, but we must show that there exist no other p that meet
the requirement. To do this is not immediately obvious but can be done quite nicely by cases. Suppose we
have p where p∈R , p6=0 then we have some multiple of i, which is clearly not in the reals. If we have a
complex, non-real p=a+bi such that a6=0 then we have a binomial to be distributed across, and the first
value of that binomial must be complex because a ∈R, thus ai∈C. So, all cases have been exhausted for
p∈C, meaning that our only candidates are ni and 0.
However, we must go back to the range of sine, which leaves us with but one possibility p=0. So, we now
go back to the main part of our problem and make the connection that p =sin2πkn =0. Now, from the
unit circle, we know that sin(φ) =0 when φ=mπ , m∈ Z. Because k and n are positive in our original,
we only consider φ=mπ , m∈N.
Now, upon investigation, we solve our equation in full and see that the nonnegative integer solutions
are, all assuming n 6= 0: k = 0 and k = 2c1c2+c2 where n = 2c2 and c1, c2 ∈ Z+ (this is a bit of a
confusing argument which could perhaps be stated more simply).This is all very general and dense, but upon requiring that we must have only real solutions for a given n, our solution nicely simplifies.
And so we can see that the only real roots of unity are±1, which are achieved often but only exclusively
for
n=1, 2
A somewhat clearer argument...
Consider the geometry of roots. We know that all nth roots of unity lie on a circle of unit radius (i.e. r=1).
Therefore, for there to be a real root it must be that z= ±1, because any other real value would be interior
or exterior to the unit disk. Now, consider the vector representation of a complex number z. Given the first
root we have some number ω0with modulus r and argument θ. Now, all other roots can be represented
by the same vector, but with the argument adjusted by some factor of θ/n, where n is the same as in the
nth root. Now, given that, because we can only achieve real numbers when the argument is kπ , k∈Z, we
have that, because all roots with n ≥3 rotate by a factor smaller than 180◦, they must attain at least some
non-real values.
31.) Discuss: A real number can have a complex nth root. Can a nonreal complex number have a real nth root?
The answer to this is a bit of yes and no. As a pseudo-example consider the famous Euler’s identity, which, I would imagine, will be discussed extensively in this course. The formula’s derivation and proof is fairly simple with methods such as Taylor series, use of Euler’s formula, which relates the exponential and polar
forms of a complex number, considerations regarding the complex exponential function ez =exp(z), where
z∈C, et cetera. Here is the identity’s statement
eiπ= −1
This simple but profound insight into mathematics also shows us an example of a number z∈C that has a
real third root−1. However, despite this number having a nice expression in terms of i, it is actually a real
number.
An actual example of this phenomenon would be to consider the number z = i, then take its ith root.
Interestingly,√i i=i1/i =i−i=eπ/2≈4.81, which, clearly, is a real number. Note here that this is actually
a multi-valued function, so we have only provided one valid answer.
That, however, is not quite what the question seems to be asking. Instead, it seems to be concerning only whole number roots given by
n √ z= √nr cos θ+2kπ n +i sin θ+2kπ n
If this be the case, then for any complex number z = a+bi , b 6= 0 we have that the magnitude is some
positive real number|z| = √a2+b2and the argument is some real number on the interval[−π/2 , π/2]
given by arctan(b/a). But, the argument cannot be zero, else the number would be real, i.e. b = 0. In
order to have a real nth root we know that the i on the sine function must be eliminated. Because the sine
function takes on only real values, the only way to eliminate the imaginary unit is to have sin(x) =0. But,
for sine to be zero we must have that θ =πk, k∈ Z, which would make the complex number real. Thus,
the only way for the root of a number to be real is if the number itself is real. There is one exception, of
course, because for any z∈C, z0=1.
This proof is a bit informal, but it demonstrates the point. In conclusion, the nth root of a nonreal complex
number z where n∈Z+cannot have a real root.
A better argument would be to argue that the reals are closed under multiplication, examining the equation z1/n =
x , x ∈R and translating it to xn =z, and finally noting that rnmust be real via closure of the group(R,·), so z is
real.
Joseph Heavner Honors Complex Analysis Assignment 2 January 25, 2015
1.5 Sets of Points in the Complex Plane
1.) Sketch the graph of
|z−4+3i| =5
.
Figure 1: Circle with radius 5 centered at 4−3i
3.) Sketch the graph of
|z+3i| =2
.
Figure 2: Circle with radius 2 centered at−3i
5.) Sketch the graph of
Re(z) =5
.
Figure 3: Vertical line at x=5
7.) Sketch the graph of
Im(z+3i) =6
Unlike the previous problems, which were obvious enough as to call for no explanation, here we will
write the equation in a more graph-able form. In particular, for z=a+bi
I(z+3i) =6
I(a−bi+3i) =6
−b+3=6
b= −3
.
Figure 4: Horizontal line at y= −3
9.) Sketch the graph of
|Re(1+iz)| =3 |R(1+i¯z)| =3 |R(1+i(x−iy))| =3 |1+y| =3 (1+y)2=9 y=2 ,−4 .
Figure 5: Horizontal lines at y=2 and y= −4
11.) Sketch the graph of Re(z2) =1 R((a+bi)2) =1 R(a2+2bi−b2) =1 a2−b2=1 . Figure 6: Hyperbola x2−y2=1
13.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected.
Re(z) < −1
a.) yes b.) no c.) yes d.) no e.) yes
Some explanation is that any neighborhood in the regionR is entirely contained (i.e. any e-ball is
entirely contained), therefore it is open. Also, not all of the boundary points ofRare contained inR,
therefore it is closed. Because any two points z0, z1∈ Rcan be connected by a collection of connected
lines, the set is connected; by definition, an open, connected set is a domain. The set is not bounded,
however, as the modulus of a point z0∈ Rcan become arbitrarily large (i.e. the set extends indefinitely
towards x= −∞).
.
Figure 7: Half-plane x< −1
15.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected.
Im(z) >3
a.) yes b.) no c.) yes d.) no e.) yes
The justification here is identical to that in (13).
.
Figure 8: Half-plane y>3
17.) Sketch the set S of the points in the complex plane satisfying the given inequality. Determine whether the set is (a) open, (b) closed, (c) a domain, (d) bounded, or (e) connected.
2<Re(z−1) <4
To make this set easier to plot, we simplify the expression describing it a bit. 2<R(z−1) <4
2<x−1 <4
3<x <5
a.) yes b.) no c.) yes d.) no e.) yes
The justification here is identical to that in (13) and (15).
.
Figure 9: 3<x<5
