Roots of Complex Numbers

11 Key excerpts on "Roots of Complex Numbers"

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  Classical Complex Analysis

  A Geometric Approach(Volume 1)

  • I-Hsiung Lin(Author)
  • 2010(Publication Date)
  • WSPC

    (Publisher)

  . . ⇒ ρ = r 1 n = | z | 1 n (as a positive real number) , ϕ = θ + 2 kπ n = 1 n (Arg z + 2 kπ ) , k = 0 , ± 1 , ± 2 , . . .. 58 Chap. 1. Complex Numbers Hence, the roots of w n = z are w k = | z | 1 n e i n ( Arg z +2 kπ ) , k = 0 , ± 1 , ± 2 , . . .. In case k < 0 or k ≥ n , w k = w m ⇔ k = pn + m for some integer p and 0 ≤ m ≤ n − 1. Hence, w n = z has only n distinct roots w k for 0 ≤ k ≤ n − 1. We summarize the above as n th roots ( n ≥ 2) of a nonzero complex number. (1) The n th roots of the unit 1 . Let ω = cos 2 π n + i sin 2 π n = e i 2 π n . Then ω n = 1 and n th roots of 1 are 1 , ω, ω 2 , . . . , ω n − 1 which form vertices of a regular n -gon inscribed in the unit circle. See Fig. 1.29 (for n = 6). (2) The n th roots of z = 0. Let θ = Arg z . Then, the n th roots of z are ζ k = | z | 1 n e i n ( θ +2 kπ ) = ζ 0 ω k , k = 0 , 1 , 2 , . . ., n − 1 , ζ 0 is usually called the principal value of n √ z . Note that ζ 0 , ζ 0 ω, . . . , ζ 0 ω n − 1 form vertices of a regular n -gon inscribed in the circle with center at 0 and radius | z | 1 n . (1.5.4) A complex number ζ , such as ω , satisfying ζ n = 1 but ζ m = 1 for any 1 ≤ m ≤ n − 1, is called a primitive n th root of 1. If ζ is a primitive n th 1 ω 2 ω 1 n ω − Fig. 1.29 Sec. 1.5. De Moivre Formula 59 root of 1, then 1 , ζ, ζ 2 , . . . , ζ n − 1 will be all the n th roots of 1. ζ = ω k is a primitive n th root if and only if 1 ≤ k ≤ n − 1 and k is relatively prime to n . Equation (1.5.3) can be extended as follows. Let n and m be integers so that n > 0 is relatively prime with | m | . Then, we define z m n =        ( z 1 n ) m , m > 0 , 1 ( z 1 n ) − m , m < 0 and z = 0 . (1.5.5) By using (1.5.4), we have z m n = | z | m n e i m ( θ +2 kπ ) n , θ = Arg z, k = 0 , 1 , 2 , . . ., n − 1 . (1.5.6) Note that, within this formula, | z | m n = n | z | m = ( n | z | ) m is chosen to be a positive number. For further discussion, see Exercise A(10). We give four examples.

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  Calculus, Volume 1

  • Tom M. Apostol(Author)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  . . , a n are arbitrary real numbers, with a n ≠ 0, has a solution among the complex numbers if n ≥ 1. Moreover, even if the coefficients a 0 , a 1 , . . . , a n are complex, a solution exists in the complex-number system. This fact is known as the fundamental theorem of algebra. † It shows that there is no need to construct numbers more general than complex numbers to solve polynomial equations with complex coefficients. 9.5 Geometric interpretation. Modulus and argument Since a complex number (x, y) is an ordered pair of real numbers, it may be represented geo- metrically by a point in the plane, or by an arrow or geometric vector from the origin to the point (x, y), as shown in Figure 9.1. In this context, the xy-plane is often referred to as the complex plane. The x-axis is called the real axis; the y-axis is the imaginary axis. It is customary to use the words complex number and point interchangeably. Thus, we refer to the point z rather than the point corresponding to the complex number z. The operations of addition and subtraction of complex numbers have a simple geometric inter- pretation. If two complex numbers z 1 and z 2 are represented by arrows from the origin to z 1 and z 2 , respectively, then the sum z 1 + z 2 is determined by the parallelogram law. The arrow from the origin to z 1 + z 2 is a diagonal of the parallelogram determined by 0, z 1 , and z 2 , as illustrated by the example in Figure 9.2. The other diagonal is related to the difference of z 1 and z 2 . The arrow from z 1 to z 2 is parallel to and equal in length to the arrow from 0 to z 2 − z 1 ; the arrow in the opposite direction, from z 2 to z 1 , is related in the same way to z 1 − z 2 . † A proof of the fundamental theorem of algebra can be found in almost any book on the theory of functions of a complex variable. For example, see K. Knopp, Theory of Functions, Dover Publications, New York, 1945, or E. Hille, Analytic Function Theory, Vol. I, Blaisdell Publishing Co., 1959.

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  Numbers and Proofs

  • Reg Allenby(Author)
  • 1997(Publication Date)
  • Butterworth-Heinemann

    (Publisher)

  NUMBERS AND PROOFS sin {§ The above is only the tip of the iceberg. We can go on to consider general functions of a complex variable, their derivatives and their integrals over contours in the complex plane. Such activity has applications in many areas of applied mathematics - even in civil engineering (where it is important in studying fluid seepage under dams). Back in pure mathematics, an estimate for the number of primes less than the integer n can be obtained via the Prime Number Theorem which may be proved by considering complex integrals and logarithms. But these are other stories. Summary Following their introduction by means of a 'problem', complex numbers were used informally by many mathematicians before it was deemed necessary to try to put them on a firmer footing. Simply defining a complex number to be 'something of the form a + ib (or a + M) where a and b are real numbers and i represents y/— Γ can lead to (temporary) difficulties - which can be eliminated by using Hamilton's definition of complex numbers using ordered pairs. Sticking to the a + ib notation we illustrated the 'trick' for simplifying a fraction of two complex numbers, and observed that the complex numbers satisfy the basic rules of arithmetic. We then stated the very important Fundamental Theorem of Algebra and deduced that, in each polynomial equation with real coefficients, the roots appear in complex conjugate pairs. The important concepts of modulus and argument were introduced via polar representation of complex numbers, and their more prominent properties - in particular the triangle inequality - were established. The neat polar form of complex multiplication leads, naturally, to De Moivre's theorem with its applications to finding roots and obtaining trigonometrical identities.

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  Trigonometry

  • Ron Larson(Author)
  • 2017(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 344 Chapter 4 Complex Numbers 2 2 n n r n π π Real axis Imaginary axis Figure 4.7 Finding nth Roots of a Complex Number For a positive integer n, the complex number z = r(cos θ + i sin θ ) has exactly n distinct nth roots given by z k = n radical.alt2r parenleft.alt4 cos θ + 2π k n + i sin θ + 2π k n parenright.alt4 where k = 0, 1, 2, . . . , n - 1. When k > n - 1, the roots begin to repeat. For example, when k = n, the angle θ + 2π n n = θ n + 2π is coterminal with θ H20862n, which is also obtained when k = 0. The formula for the nth roots of a complex number z has a geometrical interpretation, as shown in Figure 4.7. Note that the nth roots of z all have the same magnitude n radical.alt2r , so they all lie on a circle of radius n radical.alt2r with center at the origin. Furthermore, successive nth roots have arguments that differ by 2π H20862n, so the n roots are equally spaced around the circle. You have already found the sixth roots of 1 by factoring and using the Quadratic Formula. Example 2 shows how to solve the same problem with the formula for nth roots. Finding the nth Roots of a Real Number Find all sixth roots of 1. Solution First, write 1 in the trigonometric form z = 1(cos 0 + i sin 0). Then, by the nth root formula with n = 6, r = 1, and θ = 0, the roots have the form z k = 6 radical.alt21 parenleft.alt4 cos 0 + 2π k 6 + i sin 0 + 2π k 6 parenright.alt4 = cos π k 3 + i sin π k 3 . So, for k = 0, 1, 2, 3, 4, and 5, the roots are as listed below.

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  Guide to Mathematical Methods

  • John Gilbert, Camilla Jordan, David A Towers(Authors)
  • 2017(Publication Date)
  • Red Globe Press

    (Publisher)

  If a , b and c are real, we have real distinct roots, real coincident roots or a conjugate complex pair of roots depending on whether b 2 − 4 ac is positive, zero or negative. In the case where a , b and c are complex we have to find the square root of the complex number b 2 − 4 ac , for which we shall develop a technique immediately. In Section 13.9, we shall see another method, which also works for the solution of z n = a , where n is any natural number. A useful tool in solving equations with complex solutions is found by using the fact that two complex numbers are equal if and only if their real and imaginary parts are each equal. This leads to the technique of equating real and imaginary parts . Complex numbers 341 Example 13.4 Solve the equation z 2 = 3 − 4i . Put z = x + i y in the equation to obtain ( x 2 − y 2 ) + 2 xy i = 3 − 4i . Equating the real parts on the left and right gives the equation x 2 − y 2 = 3 , and equating imaginary parts gives 2 xy = − 4 . The second equation gives y = − (2 /x ), and putting this into the first equation, we obtain x 2 − 4 x 2 = 3 , which rearranges to x 4 − 3 x 2 − 4 = 0 . This is a quadratic equation in x 2 with roots x 2 = 4 or − 1, but as x is real, we must have x 2 = 4 giving x = ± 2. Since y = − (2 /x ), it follows that y = ∓ 1 and so z = ± (2 − i). We can find the square root of any complex number in this way, and hence solve any quadratic equation. Exercises: Section 13.3 1. Find all solutions in C of the equations (i) z 2 = − 8 + 6i; (ii) (1 + i) z + (2 − i) = i; (iii) | z | − ¯ z = 1 + 2i; (iv) ¯ z = z 2 . 2. Show that the equation z 2 = a = b + i c , c < 0, has solution z = ± | a | + b 2 − i | a | − b 2 . 342 Guide to Mathematical Methods 3. Solve the equation z 2 − (1 + 4i) z − (3 − i) = 0.

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  Mathematics for Information Technology

  • Alfred Basta, Stephan DeLong, Nadine Basta, , Alfred Basta, Stephan DeLong, Nadine Basta(Authors)
  • 2013(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. The Complex Numbers 309 Objectives When you have successfully completed the materials of this chapter, you will be able to: Understand the definition of complex numbers and simplify radical expressions that lead to complex numbers. Write complex numbers in rectangular, polar, and trigonometric form. Perform addition, subtraction, multiplication, division, and exponentiation of complex numbers. Solve application problems involving complex numbers. 9.1 D EFINING THE C OMPLEX N UMBERS A Reminder about Radicals The applications of the complex numbers are varied and widespread, but in order for us to develop an understanding of those applications, we must first ensure that we have a foundational knowledge of these numbers. Therefore, before proceeding further, we retreat to reminding you of the definition of the principal square root: DEFINITION: Principal of the square root: A B B A B if and only if is nonnegative and H11005 H11005 2 The value of A is referred to as the radicand of the square root, and the initial limitations on the definition require that A be nonnegative. From this definition, we can determine that 49 is equal to 7, since 7 is nonnegative and such that 49 5 7 2 . While it is true that 49 5 ( 2 7) 2 , the techni-cal requirement that B should be nonnegative compels us to choose the value of 7 as the “principal” square root of 49. Compare this problem to the search for H11002 49 . The situation is not quite the same as in the first example because the radicand value is now negative, and the definition of the principal square root would tell us that we were looking for a nonnegative number B such that 2 49 5 B 2 .

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  The Calculus Lifesaver

  All the Tools You Need to Excel at Calculus

  • Adrian Banner(Author)
  • 2009(Publication Date)
  • Princeton University Press

    (Publisher)

  Let’s set Z = z 2 , so that the equation becomes Z 2 -Z + 1 = 0. This can be solved using the quadratic formula to get Z = z 2 = 1 ± √ -3 2 = 1 2 ± i √ 3 2 . So we need to find the square roots of 1 2 + i √ 3 2 and 1 2 -i √ 3 2 . We just did the first one in the previous example, and you can repeat the steps easily enough to handle the second one. Both of these numbers have two square roots each, which work out to be √ 3 + i 2 , -√ 3 -i 2 , -√ 3 + i 2 , and √ 3 -i 2 . 610 • Complex Numbers These are the four solutions to z 4 -z 2 + 1 = 0. It follows that we can factor z 4 -z 2 + 1 as follows: z 4 -z 2 + 1 = z -√ 3 + i 2 ! z -√ 3 -i 2 ! z --√ 3 + i 2 ! z --√ 3 -i 2 ! . This is the complex factorization. To get the real factorization, we need to use a nice fact: if w is any complex number, then ( z -w )( z -¯ w ) has real coefficients when you multiply it out. Indeed, you get z 2 -( w + ¯ w ) z + w ¯ w , but it’s easy enough to see that w + ¯ w = 2Re( w ) (which is real), and we’ve already seen that w ¯ w = | w | 2 , which is also real. Anyway, notice that I have cunningly grouped the above four factors so that if we multiply out the first two, we get z -√ 3 + i 2 ! z -√ 3 -i 2 ! = z 2 - √ 3 + i 2 + √ 3 -i 2 ! z + √ 3 + i 2 ! √ 3 -i 2 ! = z 2 -√ 3 z + 1 . Similarly, you should check that multiplying out the last two factors gives z 2 + √ 3 z + 1. The conclusion is that z 4 -z 2 + 1 = ( z 2 -√ 3 z + 1)( z 2 + √ 3 z + 1) . Notice that there are no complex numbers here, yet working this out without them would have been pretty darn tricky. 28.5 Solving e z = w Now it’s time to see how to solve equations of the form e z = w for given w . It’d be nice if we could just write z = ln( w ), but this isn’t very helpful. For example, what exactly is ln( -√ 3 + i )? Let’s try to answer this question. Fortunately, solving e z = w isn’t much harder than solving z n = w ; in fact, if anything, it’s simpler. Before we see how to do this, we need to understand e z a little better.

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  Mathematical Methods in the Physical Sciences

  • Mary L. Boas(Author)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  From (10.1) you can see that this is (10.3) z 1 /n = ( re iθ ) 1 /n = r 1 /n e iθ/n = n √ r cos θ n + i sin θ n . This formula must be used with care (see Examples 2 to 4 below). Some examples will show how useful these formulas are. Example 1. [cos( π/ 10) + i sin( π/ 10)] 25 = ( e iπ/ 10 ) 25 = e 2 πi e iπ/ 2 = 1 · i = i. Section 10 Powers and Roots of Complex Numbers 65 Example 2. Find the cube roots of 8. We know that 2 is a cube root of 8, but there are also two complex cube roots of 8; let us see why. Plot the complex number 8 (that is, x = 8, y = 0) in the complex plane; the polar coordinates of the point are r = 8, and θ = 0, or 360 ◦ , 720 ◦ , 1080 ◦ , etc. (We can use either degrees or radians here; read the end of Section 3.) Now by equation (10.3), z 1 / 3 = r 1 / 3 e iθ/ 3 ; that is, to find the polar coordinates of the cube root of a number re iθ , we find the cube root of r and divide the angle by 3. Then the polar coordinates of 3 √ 8 are r = 2 , θ = 0 ◦ , 360 ◦ / 3 , 720 ◦ / 3 , 1080 ◦ / 3 · · · (10.4) = 0 ◦ , 120 ◦ , 240 ◦ , 360 ◦ · · · . We plot these points in Figure 10.1. Observe that the point (2 , 0 ◦ ) and the point (2 , 360 ◦ ) are the same. The points in (10.4) are all on a circle of radius 2 and are equally spaced 360 ◦ / 3 = 120 ◦ apart. Starting with θ = 0, if we add 120 ◦ repeatedly, we just repeat the three angles shown. Thus, there are exactly three cube roots for any number z , always on a circle of radius 3 | z | and spaced 120 ◦ apart. Now to find the values of 3 √ 8 in rectangular form, we can read them from Figure 10.1, or we can calculate them from z = r (cos θ + i sin θ ) with r = 2 and θ = 0, 120 ◦ = 2 π/ 3, 240 ◦ = 4 π/ 3. We can also use a computer to solve the equation z 3 = 8. By any of these methods we find Figure 10.1 3 √ 8 = { 2 , − 1 + i √ 3 , − 1 − i √ 3 } .

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  Maths: A Student's Survival Guide

  A Self-Help Workbook for Science and Engineering Students

  • Jenny Olive(Author)
  • 2003(Publication Date)
  • Cambridge University Press

    (Publisher)

  Suppose we take as an example the equation x 2 – 8 x + 25 = 0. It is usual to use z instead of x if we are extending the possibilities for roots by using these new numbers, so I shall rewrite this equation as z 2 – 8 z + 25 = 0. Now we see what happens if we use the quadratic formula. Using z , this formula says If az 2 + bz + c = 0 then z = – b ± b 2 – 4 ac 2 a . (Section 2.D.(d) ) Here, we get z = +8 ± 64 – 100 2 = +8 ± –36 2 = +8 ± 6 –1 2 so z = +4 ± 3 –1 giving z 1 = 4 + 3 j and z 2 = 4 – 3 j using our new number j for –1, and calling the roots z 1 and z 2 . (Try checking for yourself whether you think each of these two roots do fit the equation by substituting each of them back into the equation in turn and seeing what happens.) Notice that the two roots which we now have for this equation are each made up of two parts, the +4 which is a real number, and the 3 j which is then either added or subtracted. We can show these two roots on the same sort of diagram that we used in Figure 10.A.2(b) , with the real parts lying along the horizontal or real axis, and the imaginary parts lying along the vertical or imaginary axis. I have drawn this particular pair of roots in Figure 10.A.3. 10.A A new sort of number 425 Figure 10.A.3 Notice that they have the property that they are symmetrically placed either side of the real axis. We shall look at the special properties of pairs like this in Section 10.B.(d) . I shall now use the number 4 – 3 j shown in Figure 10.A.3 to give you some definitions. A number like 4 – 3 j is called a complex number . 4 is called its real part and –3 is called its imaginary part . The number 4 gives the measurement along the real axis. The number –3 gives the measurement along the imaginary axis. If z = 4 – 3 j then we say Re ( z ) = 4 and Im ( z ) = –3. (Notice that the imaginary part tells us how many units of j we have; it does not actually include the j .) exercise 10.

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  Complex Numbers in Geometry

  • I. M. Yaglom, Henry Booker, D. Allan Bromley, Nicholas DeClaris(Authors)
  • 2014(Publication Date)
  • Academic Press

    (Publisher)

  C H A P T E R I I Geometrical Interpretation of Complex Numbers §7. Ordinary Complex Numbers as Points of a Plane 18a The development of the theory of complex numbers is very closely connected with the geometrical interpretation of ordinary complex numbers as points of a plane, which apparently was first mentioned by the Danish surveyor K. Wessel (1745-1818) but became widely known chiefly through the works of the famous mathematicians K. F. Gauss (1777-1855) and A. Cauchy (1789-1857). This interpretation arises from the fact that the point of a plane with rectangular cartesian coordinates x and y or polar coordinates r and φ corresponds to the complex number (see Figure 1): z = x + iy = r(cos φ -- i sin ψ) Here, obviously, real numbers z = x + 0 · i = r(cos 0 + i sin 0) correspond to points of the x axis, the real axis o numbers of modulus r = 1 correspond to points of the circle S with center at O and radius 1, the unit circle. Opposite complex numbers z = x + iy and — z = —x — iy correspond to points symmet-rical about the point O (the number 0 corresponds to the origin isa Th e contents of Sections 7, 8, 13 and 14 have many points of contact with the books of R. Deaux, Introduction to the Geometry of Complex Numbers (F. Ungar Publishing Co.), 1956 and H. Schwerdtfeger, Geometry of Complex Numbers (University of Toronto, Oliver and Boyd), 1962; the latter book also contains much material which could supplement the contents of Section 11 of the present book. 26 §7. Ordinary Complex Numbers as Points of a Plane 27 »' r , - x l ...iïz FIG. 1 O); conjugate complex numbers z = x + iy = r(cos φ + i sin φ) and z = x — iy = r[cos(— φ) + ¿ sin(— φ)] correspond to points symmetrical about the line o (throughout this book we shall consider a line to mean a straight line).

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  Quantum Computing for Computer Scientists

  • Noson S. Yanofsky, Mirco A. Mannucci(Authors)
  • 2008(Publication Date)
  • Cambridge University Press

    (Publisher)

  (1.72)  Exercise 1.3.7 Divide 2 + 2i by 1 − i using both the algebraic and the geometrical method and verify that the results are the same.  You may have noticed that in Section 1.2, we have left out two important oper- ations: powers and roots. The reason was that it is much easier to deal with them in the present geometric setting than from the algebraic viewpoint. Let us begin with powers. If c = (ρ, θ ) is a complex number in polar form and n a positive integer, its nth power is just c n = (ρ n , nθ ), (1.73) because raising to the nth power is multiplying n times. Figure 1.9 shows a complex number and its first, second, and third powers. Exercise 1.3.8 Let c = 1 − i . Convert it to polar coordinates, calculate its fifth power, and revert the answers to Cartesian coordinates.  What happens when the base is a number of magnitude 1? Its powers will also have magnitude 1; thus, they will stay on the same unit circle. You can think of the various powers 1, 2, . . . as time units, and a needle moving counterclockwise at 1.3 The Geometry of Complex Numbers 23 Figure 1.9. A complex number and its square and cube. constant speed (it covers exactly θ radians per time unit, where θ is the phase of the base). Let us move on to roots. As you know already from high-school algebra, a root is a fractional power. For instance, the square root means raising the base to the power of one-half; the cube root is raising to the power of one-third; and so forth. The same holds true here, so we may take Roots of Complex Numbers: if c = (ρ, θ ) is a complex in polar form, its nth root is c 1 n =  ρ 1 n , 1 n θ  . (1.74) However, things get a bit more complicated. Remember, the phase is defined only up to multiples of 2π . Therefore, we must rewrite Equation (1.74) as c 1 n =  n √ ρ, 1 n (θ + k2π )  . (1.75) It appears that there are several roots of the same number. This fact should not surprise us: in fact, even among real numbers, roots are not always unique.

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