11.7 Polar Form of Complex Numbers


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1 11.7 Polar Form of Complex Numbers Polar Form of Complex Numbers In this section, we return to our study of complex numbers which were first introduced in Section.. Recall that a complex number is a number of the form z = a + bi where a and b are real numbers and i is the imaginary unit defined by i = 1. The number a is called the real part of z, denoted Rez, while the real number b is called the imaginary part of z, denoted Imz. From Intermediate Algebra, we know that if z = a + bi = c + di where a, b, c and d are real numbers, then a = c and b = d, which means Rez and Imz are welldefined. 1 To start off this section, we associate each complex number z = a + bi with the point a, b on the coordinate plane. In this case, the xaxis is relabeled as the real axis, which corresponds to the real number line as usual, and the yaxis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. Imaginary Axis i i, z = + i i i, 0 z = Real Axis i i i 0, z = i i The Complex Plane Since the ordered pair a, b gives the rectangular coordinates associated with the complex number z = a + bi, the expression z = a + bi is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates r, θ. Although it is not a straightforward as the definitions of Rez and Imz, we can still give r and θ special names in relation to z. Definition 11.. The Modulus and Argument of Complex Numbers: Let z = a + bi be a complex number with a = Rez and b = Imz. Let r, θ be a polar representation of the point with rectangular coordinates a, b where r 0. The modulus of z, denoted z, is defined by z = r. The angle θ is an argument of z. The set of all arguments of z is denoted argz. If z 0 and < θ, then θ is the principal argument of z, written θ = Argz. 1 Welldefined means that no matter how we express z, the number Rez is always the same, and the number Imz is always the same. In other words, Re and Im are functions of complex numbers.
2 990 Applications of Trigonometry Some remarks about Definition 11. are in order. We know from Section 11. that every point in the plane has infinitely many polar coordinate representations r, θ which means it s worth our time to make sure the quantities modulus, argument and principal argument are welldefined. Concerning the modulus, if z = 0 then the point associated with z is the origin. In this case, the only rvalue which can be used here is r = 0. Hence for z = 0, z = 0 is welldefined. If z 0, then the point associated with z is not the origin, and there are two possibilities for r: one positive and one negative. However, we stipulated r 0 in our definition so this pins down the value of z to one and only one number. Thus the modulus is welldefined in this case, too. Even with the requirement r 0, there are infinitely many angles θ which can be used in a polar representation of a point r, θ. If z 0 then the point in question is not the origin, so all of these angles θ are coterminal. Since coterminal angles are exactly radians apart, we are guaranteed that only one of them lies in the interval, ], and this angle is what we call the principal argument of z, Argz. In fact, the set argz of all arguments of z can be described using setbuilder notation as argz = {Argz + k k is an integer}. Note that since argz is a set, we will write θ argz to mean θ is in the set of arguments of z. If z = 0 then the point in question is the origin, which we know can be represented in polar coordinates as 0, θ for any angle θ. In this case, we have arg0 =, and since there is no one value of θ which lies, ], we leave Arg0 undefined. It is time for an example. Example For each of the following complex numbers find Rez, Imz, z, argz and Argz. Plot z in the complex plane. 1. z = i. z = + i. z = i. z = 117 Solution. 1. For z = i = + 1i, we have Rez = and Imz = 1. To find z, argz and Argz, we need to find a polar representation r, θ with r 0 for the point P, 1 associated with z. We know r = + 1 =, so r = ±. Since we require r 0, we choose r =, so z =. Next, we find a corresponding angle θ. Since r > 0 and P lies in Quadrant IV, θ is a Quadrant IV angle. We know tanθ = 1 =, so θ = 6 + k for integers k. Hence, argz = { 6 + k k is an integer}. Of these values, only θ = 6 satisfies the requirement that < θ, hence Argz = 6.. The complex number z = + i has Rez =, Imz =, and is associated with the point P,. Our next task is to find a polar representation r, θ for P where r 0. Running through the usual calculations gives r = 5, so z = 5. To find θ, we get tanθ =, and since r > 0 and P lies in Quadrant II, we know θ is a Quadrant II angle. We find θ = + arctan + k, or, more succinctly θ = arctan + k for integers k. Hence argz = { arctan + k k is an integer}. Only θ = arctan satisfies the requirement < θ, so Argz = arctan. In case you re wondering, the use of the absolute value notation z for modulus will be explained shortly. Recall the symbol being used here,, is the mathematical symbol which denotes membership in a set. If we had Calculus, we would regard Arg0 as an indeterminate form. But we don t, so we won t.
3 11.7 Polar Form of Complex Numbers 991. We rewrite z = i as z = 0 + i to find Rez = 0 and Imz =. The point in the plane which corresponds to z is 0, and while we could go through the usual calculations to find the required polar form of this point, we can almost see the answer. The point 0, lies units away from the origin on the positive yaxis. Hence, r = z = and θ = + k for integers k. We get argz = { + k k is an integer} and Argz =.. As in the previous problem, we write z = 117 = i so Rez = 117 and Imz = 0. The number z = 117 corresponds to the point 117, 0, and this is another instance where we can determine the polar form by eye. The point 117, 0 is 117 units away from the origin along the negative xaxis. Hence, r = z = 117 and θ = + = k + 1k for integers k. We have argz = {k + 1 k is an integer}. Only one of these values, θ =, just barely lies in the interval, ] which means and Argz =. We plot z along with the other numbers in this example below. Imaginary Axis z = + i z = 117 i i i i z = i i z = i Real Axis Now that we ve had some practice computing the modulus and argument of some complex numbers, it is time to explore their properties. We have the following theorem. Theorem Properties of the Modulus: Let z and w be complex numbers. z is the distance from z to 0 in the complex plane z 0 and z = 0 if and only if z = 0 z = Rez + Imz Product Rule: zw = z w Power Rule: z n = z n for all natural numbers, n Quotient Rule: z = z w w, provided w 0 To prove the first three properties in Theorem 11.1, suppose z = a + bi where a and b are real numbers. To determine z, we find a polar representation r, θ with r 0 for the point a, b. From Section 11., we know r = a + b so that r = ± a + b. Since we require r 0, then it must be that r = a + b, which means z = a + b. Using the distance formula, we find the distance
4 99 Applications of Trigonometry from 0, 0 to a, b is also a + b, establishing the first property. 5 For the second property, note that since z is a distance, z 0. Furthermore, z = 0 if and only if the distance from z to 0 is 0, and the latter happens if and only if z = 0, which is what we were asked to show. 6 For the third property, we note that since a = Rez and b = Imz, z = a + b = Rez + Imz. To prove the product rule, suppose z = a + bi and w = c + di for real numbers a, b, c and d. Then zw = a + bic + di. After the usual arithmetic 7 we get zw = ac bd + ad + bci. Therefore, zw = ac bd + ad + bc = a c abcd + b d + a d + abcd + b c Expand = a c + a d + b c + b d Rearrange terms = a c + d + b c + d Factor = a + b c + d Factor = a + b c + d Product Rule for Radicals = z w Definition of z and w Hence zw = z w as required. Now that the Product Rule has been established, we use it and the Principle of Mathematical Induction 8 to prove the power rule. Let P n be the statement z n = z n. Then P 1 is true since z 1 = z = z 1. Next, assume P k is true. That is, assume z k = z k for some k 1. Our job is to show that P k + 1 is true, namely z k+1 = z k+1. As is customary with induction proofs, we first try to reduce the problem in such a way as to use the Induction Hypothesis. z k+1 = z k z Properties of Exponents = z k z Product Rule = z k z Induction Hypothesis = z k+1 Properties of Exponents Hence, P k + 1 is true, which means z n = z n is true for all natural numbers n. Like the Power Rule, the Quotient Rule can also be established with the help of the Product Rule. We assume w 0 so w 0 and we get z = 1 w z w = z 1 w Product Rule. 5 Since the absolute value x of a real number x can be viewed as the distance from x to 0 on the number line, this first property justifies the notation z for modulus. We leave it to the reader to show that if z is real, then the definition of modulus coincides with absolute value so the notation z is unambiguous. 6 This may be considered by some to be a bit of a cheat, so we work through the underlying Algebra to see this is true. We know z = 0 if and only if a + b = 0 if and only if a + b = 0, which is true if and only if a = b = 0. The latter happens if and only if z = a + bi = 0. There. 7 See Example..1 in Section. for a review of complex number arithmetic. 8 See Section 9. for a review of this technique.
5 11.7 Polar Form of Complex Numbers 99 Hence, the proof really boils down to showing 1 w = 1 w. This is left as an exercise. Next, we characterize the argument of a complex number in terms of its real and imaginary parts. Theorem Properties of the Argument: Let z be a complex number. If Rez 0 and θ argz, then tanθ = Imz Rez. If Rez = 0 and Imz > 0, then argz = { + k k is an integer}. If Rez = 0 and Imz < 0, then argz = { + k k is an integer}. If Rez = Imz = 0, then z = 0 and argz =,. To prove Theorem 11.15, suppose z = a+bi for real numbers a and b. By definition, a = Rez and b = Imz, so the point associated with z is a, b = Rez, Imz. From Section 11., we know that if r, θ is a polar representation for Rez, Imz, then tanθ = Imz Rez, provided Rez 0. If Rez = 0 and Imz > 0, then z lies on the positive imaginary axis. Since we take r > 0, we have that θ is coterminal with, and the result follows. If Rez = 0 and Imz < 0, then z lies on the negative imaginary axis, and a similar argument shows θ is coterminal with. The last property in the theorem was already discussed in the remarks following Definition 11.. Our next goal is to completely marry the Geometry and the Algebra of the complex numbers. To that end, consider the figure below. Imaginary Axis a, b z = a + bi r, θ bi z = a + b = r 0 θ argz a Real Axis Polar coordinates, r, θ associated with z = a + bi with r 0. We know from Theorem 11.7 that a = r cosθ and b = r sinθ. Making these substitutions for a and b gives z = a + bi = r cosθ + r sinθi = r [cosθ + i sinθ]. The expression cosθ + i sinθ is abbreviated cisθ so we can write z = rcisθ. Since r = z and θ argz, we get Definition 11.. A Polar Form of a Complex Number: Suppose z is a complex number and θ argz. The expression: is called a polar form for z. z cisθ = z [cosθ + i sinθ]
6 99 Applications of Trigonometry Since there are infinitely many choices for θ argz, there infinitely many polar forms for z, so we used the indefinite article a in Definition 11.. It is time for an example. Example Find the rectangular form of the following complex numbers. Find Rez and Imz. a z = cis b z = cis c z = cis0 d z = cis. Use the results from Example to find a polar form of the following complex numbers. a z = i b z = + i c z = i d z = 117 Solution. 1. The key to this problem is to write out cisθ as cosθ + i sinθ. a By definition, z = cis [ = cos + i sin ]. After some simplifying, we get z = + i, so that Rez = and Imz =. = [ cos b Expanding, we get z = cis z = i, so Rez = = Imz. + i sin ]. From this, we find c We get z = cis0 = [cos0 + i sin0] =. Writing = + 0i, we get Rez = and Imz = 0, which makes sense seeing as is a real number. d Lastly, we have z = cis = cos + i sin = i. Since i = 0 + 1i, we get Rez = 0 and Imz = 1. Since i is called the imaginary unit, these answers make perfect sense.. To write a polar form of a complex number z, we need two pieces of information: the modulus z and an argument not necessarily the principal argument of z. We shamelessly mine our solution to Example to find what we need. a For z = i, z = and θ = 6, so z = cis 6. We can check our answer by converting it back to rectangular form to see that it simplifies to z = i. b For z = + i, z = 5 and θ = arctan. Hence, z = 5cis arctan. It is a good exercise to actually show that this polar form reduces to z = + i. c For z = i, z = and θ =. In this case, z = cis. This can be checked geometrically. Head out units from 0 along the positive real axis. Rotating radians counterclockwise lands you exactly units above 0 on the imaginary axis at z = i. d Last but not least, for z = 117, z = 117 and θ =. We get z = 117cis. As with the previous problem, our answer is easily checked geometrically.
7 11.7 Polar Form of Complex Numbers 995 The following theorem summarizes the advantages of working with complex numbers in polar form. Theorem Products, Powers and Quotients Complex Numbers in Polar Form: Suppose z and w are complex numbers with polar forms z = z cisα and w = w cisβ. Then Product Rule: zw = z w cisα + β Power Rule DeMoivre s Theorem : z n = z n cisnθ for every natural number n Quotient Rule: z w = z cisα β, provided w 0 w The proof of Theorem requires a healthy mix of definition, arithmetic and identities. We first start with the product rule. zw = [ z cisα] [ w cisβ] = z w [cosα + i sinα] [cosβ + i sinβ] We now focus on the quantity in brackets on the right hand side of the equation. [cosα + i sinα] [cosβ + i sinβ] = cosα cosβ + i cosα sinβ + i sinα cosβ + i sinα sinβ = cosα cosβ + i sinα sinβ Rearranging terms + i sinα cosβ + i cosα sinβ = cosα cosβ sinα sinβ Since i = 1 + i sinα cosβ + cosα sinβ Factor out i = cosα + β + i sinα + β Sum identities = cisα + β Definition of cis Putting this together with our earlier work, we get zw = z w cisα + β, as required. Moving right along, we next take aim at the Power Rule, better known as DeMoivre s Theorem. 9 We proceed by induction on n. Let P n be the sentence z n = z n cisnθ. Then P 1 is true, since z 1 = z = z cisθ = z 1 cis1 θ. We now assume P k is true, that is, we assume z k = z k ciskθ for some k 1. Our goal is to show that P k + 1 is true, or that z k+1 = z k+1 cisk + 1θ. We have z k+1 = z k z Properties of Exponents = z k ciskθ z cisθ Induction Hypothesis = z k z ciskθ + θ Product Rule = z k+1 cisk + 1θ 9 Compare this proof with the proof of the Power Rule in Theorem 11.1.
8 996 Applications of Trigonometry Hence, assuming P k is true, we have that P k + 1 is true, so by the Principle of Mathematical Induction, z n = z n cisnθ for all natural numbers n. The last property in Theorem to prove is the quotient rule. Assuming w 0 we have z w = z cisα w cisβ z cosα + i sinα = w cosβ + i sinβ Next, we multiply both the numerator and denominator of the right hand side by cosβ i sinβ which is the complex conjugate of cosβ + i sinβ to get z w = z cosα + i sinα cosβ i sinβ w cosβ + i sinβ cosβ i sinβ If we let the numerator be N = [cosα + i sinα] [cosβ i sinβ] and simplify we get N = [cosα + i sinα] [cosβ i sinβ] = cosα cosβ i cosα sinβ + i sinα cosβ i sinα sinβ Expand = [cosα cosβ + sinα sinβ] + i [sinα cosβ cosα sinβ] Rearrange and Factor = cosα β + i sinα β Difference Identities = cisα β Definition of cis If we call the denominator D then we get D = [cosβ + i sinβ] [cosβ i sinβ] = cos β i cosβ sinβ + i cosβ sinβ i sin β Expand = cos β i sin β Simplify = cos β + sin β Again, i = 1 = 1 Pythagorean Identity Putting it all together, we get z w = = z cosα + i sinα w cosβ + i sinβ z cisα β w 1 = z cisα β w cosβ i sinβ cosβ i sinβ and we are done. The next example makes good use of Theorem
9 11.7 Polar Form of Complex Numbers 997 Example Let z = + i and w = 1 + i. Use Theorem to find the following. 1. zw. w 5. Write your final answers in rectangular form. Solution. In order to use Theorem 11.16, we need to write z and w in polar form. For z = +i, we find z = + = 16 =. If θ argz, we know tanθ = Imz Rez = =. Since z lies in Quadrant I, we have θ = 6 + k for integers k. Hence, z = cis 6. For w = 1 + i, we have w = 1 + =. For an argument θ of w, we have tanθ = 1 =. Since w lies in Quadrant II, θ = + k for integers k and w = cis. We can now proceed. 1. We get zw = cis 6 cis = 8cis 6 + = 8cis 5 [ 6 = 8 cos i sin 5 ] 6. After simplifying, we get zw = + i.. We use DeMoivre s Theorem which yields w 5 = [ cis ] 5 = 5 cis 5 = cis 10. Since 10 is coterminal with, we get w5 = [ cos + i sin ] = 16 16i.. Last, but not least, we have z w = cis 6 cis = cis 6 = cis. Since is a quadrantal angle, we can see the rectangular form by moving out units along the positive real axis, then rotating radians clockwise to arrive at the point units below 0 on the imaginary axis. The long and short of it is that z w = i. Some remarks are in order. First, the reader may not be sold on using the polar form of complex numbers to multiply complex numbers especially if they aren t given in polar form to begin with. Indeed, a lot of work was needed to convert the numbers z and w in Example into polar form, compute their product, and convert back to rectangular form certainly more work than is required to multiply out zw = + i 1 + i the oldfashioned way. However, Theorem pays huge dividends when computing powers of complex numbers. Consider how we computed w 5 above and compare that to using the Binomial Theorem, Theorem 9., to accomplish the same feat by expanding 1 + i 5. Division is tricky in the best of times, and we saved ourselves a lot of time and effort using Theorem to find and simplify z w using their polar forms as opposed to starting with +i 1+i, rationalizing the denominator, and so forth. There is geometric reason for studying these polar forms and we would be derelict in our duties if we did not mention the Geometry hidden in Theorem Take the product rule, for instance. If z = z cisα and w = w cisβ, the formula zw = z w cisα + β can be viewed geometrically as a two step process. The multiplication of z by w can be interpreted as magnifying 10 the distance z from z to 0, by the factor w. Adding the argument of w to the argument of z can be interpreted geometrically as a rotation of β radians counterclockwise. 11 Focusing on z and w from Example 10 Assuming w > Assuming β > 0. z w
10 998 Applications of Trigonometry 11.7., we can arrive at the product zw by plotting z, doubling its distance from 0 since w =, and rotating radians counterclockwise. The sequence of diagrams below attempts to describe this process geometrically. Imaginary Axis Imaginary Axis 6i 5i i i i i z = cis ` 6 z w = 8cis ` 6 zw = 8cis ` 6 + 6i 5i i i i i z w = 8cis ` Real Axis Real Axis Multiplying z by w =. Rotating counterclockwise by Argw = radians. Visualizing zw for z = cis ` ` 6 and w = cis. We may also visualize division similarly. Here, the formula z w = z w cisα β may be interpreted as shrinking 1 the distance from 0 to z by the factor w, followed up by a clockwise 1 rotation of β radians. In the case of z and w from Example 11.7., we arrive at z w by first halving the distance from 0 to z, then rotating clockwise radians. Imaginary Axis Imaginary Axis i i 1 z = cis ` w 6 i i z = cis ` 6 1 z = cis ` w 6 i 0 1 Real Axis 0 1 Real Axis i zw = cis ` 6 Dividing z by w =. Rotating clockwise by Argw = radians. Visualizing z w for z = cis ` ` 6 and w = cis. Our last goal of the section is to reverse DeMoivre s Theorem to extract roots of complex numbers. Definition 11.. Let z and w be complex numbers. If there is a natural number n such that w n = z, then w is an n th root of z. Unlike Definition 5. in Section 5., we do not specify one particular prinicpal n th root, hence the use of the indefinite article an as in an n th root of z. Using this definition, both and are 1 Again, assuming w > 1. 1 Again, assuming β > 0.
11 11.7 Polar Form of Complex Numbers 999 square roots of 16, while 16 means the principal square root of 16 as in 16 =. Suppose we wish to find all complex third cube roots of 8. Algebraically, we are trying to solve w = 8. We know that there is only one real solution to this equation, namely w = 8 =, but if we take the time to rewrite this equation as w 8 = 0 and factor, we get w w + w + = 0. The quadratic factor gives two more cube roots w = 1 ± i, for a total of three cube roots of 8. In accordance with Theorem.1, since the degree of pw = w 8 is three, there are three complex zeros, counting multiplicity. Since we have found three distinct zeros, we know these are all of the zeros, so there are exactly three distinct cube roots of 8. Let us now solve this same problem using the machinery developed in this section. To do so, we express z = 8 in polar form. Since z = 8 lies 8 units away on the positive real axis, we get z = 8cis0. If we let w = w cisα be a polar form of w, the equation w = 8 becomes w = 8 w cisα = 8cis0 w cisα = 8cis0 DeMoivre s Theorem The complex number on the left hand side of the equation corresponds to the point with polar coordinates w, α, while the complex number on the right hand side corresponds to the point with polar coordinates 8, 0. Since w 0, so is w, which means w, α and 8, 0 are two polar representations corresponding to the same complex number, both with positive r values. From Section 11., we know w = 8 and α = 0 + k for integers k. Since w is a real number, we solve w = 8 by extracting the principal cube root to get w = 8 =. As for α, we get α = k for integers k. This produces three distinct points with polar coordinates corresponding to k = 0, 1 and : specifically, 0,, and,. These correspond to the complex numbers w 0 = cis0, w 1 = cis and w = cis, respectively. Writing these out in rectangular form yields w 0 =, w 1 = 1 + i and w = 1 i. While this process seems a tad more involved than our previous factoring approach, this procedure can be generalized to find, for example, all of the fifth roots of. Try using Chapter techniques on that! If we start with a generic complex number in polar form z = z cisθ and solve w n = z in the same manner as above, we arrive at the following theorem. Theorem The n th roots of a Complex Number: Let z 0 be a complex number with polar form z = rcisθ. For each natural number n, z has n distinct n th roots, which we denote by w 0, w 1,..., w n 1, and they are given by the formula w k = n θ rcis n + n k The proof of Theorem breaks into to two parts: first, showing that each w k is an n th root, and second, showing that the set {w k k = 0, 1,..., n 1} consists of n different complex numbers. To show w k is an n th root of z, we use DeMoivre s Theorem to show w k n = z.
12 1000 Applications of Trigonometry w k n = n rcis θ n + n k n = n r n cis n [ θ n + n k] = rcis θ + k DeMoivre s Theorem Since k is a whole number, cosθ + k = cosθ and sinθ + k = sinθ. Hence, it follows that cisθ + k = cisθ, so w k n = rcisθ = z, as required. To show that the formula in Theorem generates n distinct numbers, we assume n or else there is nothing to prove and note that the modulus of each of the w k is the same, namely n r. Therefore, the only way any two of these polar forms correspond to the same number is if their arguments are coterminal that is, if the arguments differ by an integer multiple of. Suppose k and j are whole numbers between 0 and n 1, inclusive, with k j. Since k and j are different, let s assume for the sake of argument k j n. For this to be an integer multiple of, that k > j. Then θ n + n k θ n + n j = k j must be a multiple of n. But because of the restrictions on k and j, 0 < k j n 1. Think this through. Hence, k j is a positive number less than n, so it cannot be a multiple of n. As a result, w k and w j are different complex numbers, and we are done. By Theorem.1, we know there at most n distinct solutions to w n = z, and we have just found all of them. We illustrate Theorem in the next example. Example Use Theorem to find the following: 1. both square roots of z = + i. the four fourth roots of z = 16. the three cube roots of z = + i. the five fifth roots of z = 1. Solution. 1. We start by writing z = + i = cis. To use Theorem 11.17, we identify r =, θ = and n =. We know that z has two square roots, and in keeping with the notation in Theorem 11.17, we ll call them w 0 and w 1. We get w 0 = cis / + 0 = cis and w 1 = cis / + 1 = cis. In rectangular form, the two square roots of z are w 0 = 1 + i and w 1 = 1 i. We can check our answers by squaring them and showing that we get z = + i.. Proceeding as above, we get z = 16 = 16cis. With r = 16, θ = and n =, we get the four fourth roots of z to be w 0 = 16cis + 0 = cis, w1 = 16cis + 1 = cis, w = 16cis + = cis 5 and w = 16cis + = cis 7. Converting these to rectangular form gives w 0 = +i, w 1 = +i, w = i and w = i.
13 11.7 Polar Form of Complex Numbers For z = +i, we have z = cis. With r =, θ = and n = the usual computations yield w 0 = cis 1, w1 = cis 9 1 = cis and w = cis If we were to convert these to rectangular form, we would need to use either the Sum and Difference Identities in Theorem or the HalfAngle Identities in Theorem to evaluate w 0 and w. Since we are not explicitly told to do so, we leave this as a good, but messy, exercise.. To find the five fifth roots of 1, we write 1 = 1cis0. We have r = 1, θ = 0 and n = 5. Since 5 1 = 1, the roots are w 0 = cis0 = 1, w 1 = cis 5, w = cis 5, w = cis 6 5 and w = cis 8 5. The situation here is even graver than in the previous example, since we have not developed any identities to help us determine the cosine or sine of 5. At this stage, we could approximate our answers using a calculator, and we leave this as an exercise. Now that we have done some computations using Theorem 11.17, we take a step back to look at things geometrically. Essentially, Theorem says that to find the n th roots of a complex number, we first take the n th root of the modulus and divide the argument by n. This gives the first root w 0. Each succeessive root is found by adding n to the argument, which amounts to rotating w 0 by n radians. This results in n roots, spaced equally around the complex plane. As an example of this, we plot our answers to number in Example below. Imaginary Axis i w 1 i w Real Axis i w w i The four fourth roots of z = 16 equally spaced = around the plane. We have only glimpsed at the beauty of the complex numbers in this section. The complex plane is without a doubt one of the most important mathematical constructs ever devised. Coupled with Calculus, it is the venue for incredibly important Science and Engineering applications. 1 For now, the following exercises will have to suffice. 1 For more on this, see the beautifully written epilogue to Section. found on page 9.
14 100 Applications of Trigonometry Exercises In Exercises 10, find a polar representation for the complex number z and then identify Rez, Imz, z, argz and Argz. 1. z = 9 + 9i. z = 5 + 5i. z = 6i. z = +i 5. z = 6 + 6i 6. z = 7. z = 1 i 8. z = i 9. z = 5i 10. z = i 11. z = 6 1. z = i 7 1. z = + i 1. z = + i 15. z = 7 + i 16. z = + 6i 17. z = 1 5i 18. z = 5 i 19. z = i 0. z = 1 i In Exercises 10, find the rectangular form of the given complex number. Use whatever identities are necessary to find the exact values. 1. z = 6cis0. z = cis. z = 7 cis. z = cis 6 5. z = cis 9. z = 7cis. z = 8cis 1 5. z = 5cis arctan 6. z = 6cis 0. z = 1cis 7. z = 9cis 8. z = cis 1. z = 1 7 cis. z = 1cis 7. z = cis 8 6. z = 10cis arctan 7. z = 15cis arctan 8. z = arctan 9. z = 50cis 7 arctan 1 0. z = 1 5 cis + arctan 1 For Exercises 15, use z = + i and w = i to compute the quantity. Express your answers in polar form using the principal argument. 1. zw. z w 5. w 6. z 5 w 7. z w 8.. w z. z z w
15 11.7 Polar Form of Complex Numbers w z 50. z w 51. w z 5. w z 6 In Exercises 56, use DeMoivre s Theorem to find the indicated power of the given complex number. Express your final answers in rectangular form i i i i i 1 6 i 59. i i i 6. + i 5 6. i i 8 In Exercises 6576, find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. 65. the two square roots of z = i 66. the two square roots of z = 5i 67. the two square roots of z = 1 + i 68. the two square roots of 5 5 i 69. the three cube roots of z = the three cube roots of z = the three cube roots of z = i 7. the three cube roots of z = 8i 7. the four fourth roots of z = the four fourth roots of z = the six sixth roots of z = the six sixth roots of z = Use the Sum and Difference Identities in Theorem or the Half Angle Identities in Theorem to express the three cube roots of z = + i in rectangular form. See Example 11.7., number. 78. Use a calculator to approximate the five fifth roots of 1. See Example 11.7., number. 79. According to Theorem.16 in Section., the polynomial px = x + can be factored into the product linear and irreducible quadratic factors. In Exercise 8 in Section 8.7, we showed you how to factor this polynomial into the product of two irreducible quadratic factors using a system of nonlinear equations. Now that we can compute the complex fourth roots of directly, we can simply apply the Complex Factorization Theorem, Theorem.1, to obtain the linear factorization px = x 1 + ix 1 ix 1 + ix 1 i. By multiplying the first two factors together and then the second two factors together, thus pairing up the complex conjugate pairs of zeros Theorem.15 told us we d get, we have that px = x x + x + x +. Use the 1 complex 1 th roots of 096 to factor px = x into a product of linear and irreducible quadratic factors.
16 100 Applications of Trigonometry 80. Complete the proof of Theorem 11.1 by showing that if w 0 than 1 = Recall from Section. that given a complex number z = a+bi its complex conjugate, denoted z, is given by z = a bi. a Prove that z = z. b Prove that z = zz c Show that Rez = z + z and Imz = z z i d Show that if θ argz then θ arg z. Interpret this result geometrically. e Is it always true that Arg z = Argz? 8. Given any natural number n, the n complex n th roots of the number z = 1 are called the n th Roots of Unity. In the following exercises, assume that n is a fixed, but arbitrary, natural number such that n. a Show that w = 1 is an n th root of unity. b Show that if both w j and w k are n th roots of unity then so is their product w j w k. c Show that if w j is an n th root of unity then there exists another n th root of unity w j such that w j w j = 1. Hint: If w j = cisθ let w j = cis θ. You ll need to verify that w j = cis θ is indeed an n th root of unity. 8. Another way to express the polar form of a complex number is to use the exponential function. For real numbers t, Euler s Formula defines e it = cost + i sint. a Use Theorem to show that e ix e iy = e ix+y for all real numbers x and y. b Use Theorem to show that e ix n = e inx for any real number x and any natural number n. c Use Theorem to show that eix e iy = eix y for all real numbers x and y. d If z = rcisθ is the polar form of z, show that z = re it where θ = t radians. e Show that e i + 1 = 0. This famous equation relates the five most important constants in all of Mathematics with the three most fundamental operations in Mathematics. f Show that cost = eit + e it w w. and that sint = eit e it for all real numbers t. i
17 11.7 Polar Form of Complex Numbers Answers 1. z = 9 + 9i = 9 cis, Rez = 9, Imz = 9, z = 9 argz = { + k k is an integer} and Argz =.. z = 5 + 5i = 10cis, Rez = 5, Imz = 5, z = 10 argz = { + k k is an integer} and Argz =.. z = 6i = 6cis, Rez = 0, Imz = 6, z = 6 argz = { + k k is an integer} and Argz =.. z = + i = 6cis, Rez =, Imz =, z = 6 argz = { + k k is an integer} and Argz =. 5. z = 6 + 6i = 1cis 5 6, Rez = 6, Imz = 6, z = 1 argz = { k k is an integer} and Argz = z = = cis, Rez =, Imz = 0, z = argz = {k + 1 k is an integer} and Argz =. 7. z = 1 i = cis 7 6, Rez =, Imz = 1, z = 1 argz = { k k is an integer} and Argz = 5 6., Rez =, Imz =, z = 8. z = i = cis 5 argz = { 5 + k k is an integer} and Argz =., Rez = 0, Imz = 5, z = 5 9. z = 5i = 5cis argz = { + k k is an integer} and Argz =., Rez =, Imz =, z = 10. z = i = cis 7 argz = { 7 + k k is an integer} and Argz =. 11. z = 6 = 6cis 0, Rez = 6, Imz = 0, z = 6 argz = {k k is an integer} and Argz = z = i 7 = 7cis, Rez = 0, Imz = 7, z = 7 argz = { + k k is an integer} and Argz =. 1. z = + i = 5cis arctan, Rez =, Imz =, z = 5 argz = { arctan + k k is an integer } and Argz = arctan.
18 1006 Applications of Trigonometry 1. z = + i = cis argz = { arctan, Rez =, Imz = 1, z = } and Argz = arctan arctan + k k is an integer 15. z = 7 + i = 5cis arctan 7, Rez = 7, Imz =, z = 5 argz = { arctan } 7 + k k is an integer and Argz = arctan z = + 6i = 10cis arctan, Rez =, Imz = 6, z = 10 argz = { arctan + k k is an integer} and Argz = arctan. 17. z = 1 5i = 1cis + arctan 5 1, Rez = 1, Imz = 5, z = 1 argz = { + arctan } k k is an integer and Argz = arctan z = 5 i = 9cis + arctan 5, Rez = 5, Imz =, z = 9 argz = { + arctan 5 + k k is an integer } and Argz = arctan z = i = 5cis arctan 1, Rez =, Imz =, z = 5 argz = { arctan 1 + k k is an integer } and Argz = arctan 1 = arctan z = 1 i = 10cis arctan, Rez = 1, Imz =, z = 10 argz = {arctan + k k is an integer} and Argz = arctan = arctan. 1. z = 6cis0 = 6. z = cis 6 = + i. z = 7 cis = 7 + 7i. z = cis = i 5. z = cis = + i 6. z = 6cis = + i 7. z = 9cis = 9 8. z = cis = i 9. z = 7cis = 7 7 i 0. z = 1cis = i z = 1 cis 7 = i. z = 1cis = 6 6i. z = 8cis 1 = + + i. z = cis 7 8 = + + i 5. z = 5cis arctan = + i 6. z = 10cis arctan 1 = + i 7. z = 15cis arctan = 5 6i 5 8. z = cis arctan = 1 i 9. z = 50cis arctan 7 = 8 + 1i 0. z = 1 cis + arctan 5 1 = 6 1 5i 6
19 11.7 Polar Form of Complex Numbers 1007 In Exercises 15, we have that z = + i = cis 5 6 and w = i = 6cis so we get the following. 1. zw = 18cis 7 1. z = 81cis z. w = 1 cis w = 16cis 7. z w z = 97cis0 8. w = cis 1 w. z = cis z 5 w = 878cis 9. w z = cis 1 z 50. = w cis 51. w = z cis 5. w 6 z = 6cis 5. + i = 6 5. i = 8i i = i = 8 + 8i i 6 = i i = i = 7 7 i i = i i = i i 6. i 5 = 16 16i 6. 1 i 8 = Since z = i = cis we have w 0 = cis = + i w1 = cis 5 = i 66. Since z = 5i = 5cis we have w 0 = 5cis = i w 1 = 5cis 7 = 5 5 i 67. Since z = 1 + i = cis we have w 0 = cis 6 = 6 + i w 1 = cis 7 6 = 6 i 68. Since z = 5 5 i = 5cis 5 we have w 0 = 5cis 5 6 = i w 1 = 5cis 11 6 = 15 5 i 69. Since z = 6 = 6cis 0 we have w 0 = cis 0 = w 1 = cis = + i w = cis = i
20 1008 Applications of Trigonometry 70. Since z = 15 = 15cis we have w 0 = 5cis = i w 1 = 5cis = 5 w = 5cis 5 = 5 5 i 71. Since z = i = cis we have w 0 = cis 6 = + 1 i w 1 = cis 5 6 = + 1 i w = cis = i 7. Since z = 8i = 8cis we have w 0 = cis = i w1 = cis Since z = 16 = 16cis 0 we have = i w = cis 11 6 = i w 0 = cis 0 = w = cis = w 1 = cis = i w = cis = i 7. Since z = 81 = 81cis we have w 0 = cis = + i w 1 = cis = + w = cis 5 = i w = cis 7 i = i 75. Since z = 6 = 6cis0 we have w 0 = cis0 = w 1 = cis = 1 + i w = cis = 1 + i w = cis = w = cis = 1 i w5 = cis = 1 i 76. Since z = 79 = 79cis we have w 0 = cis 6 = + i w 1 = cis w = cis 7 6 = i w = cis = i w = cis 5 6 = + i = i w5 = cis 11 6 = i 77. Note: In the answers for w 0 and w the first rectangular form comes from applying the appropriate Sum or Difference Identity 1 = 17 and 1 = +, respectively and the second comes from using the HalfAngle Identities. w 0 = cis = + i = + + i w 1 = cis = + i w = cis 17 1 = 6 + i 6 = + + i
21 11.7 Polar Form of Complex Numbers w 0 = cis0 = 1 w 1 = cis i w = cis i w = cis i w = cis i 79. px = x = x x+x +x x+x +x+x x+x + +
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