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 , −−→z
1z3=z3−z1 , −−→z
2z3=z3−z2
4
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 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
17.) Describe the set of points z in the complex plane that satisfy Re((1+i)z−1) =0
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.
6
37.) Under what circumstance does|z1+z2| = |z1| + |z2|? So, we first square both sides as follows
(a+x)2+ (b+y)2=2p
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+.
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 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|
8