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 i sin(7π/4)and simplifies to our final answer
√4
19.)
(a) Show that the n nth roots of unity are given by (1)1/n=cos2πk
n +i sin2πk
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
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±p22−4(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
−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.
18
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.
19
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
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 =sin
2πk
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.
20
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 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.
21
Joseph Heavner Honors Complex Analysis Assignment 2 January 25, 2015