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Introduction De Moivre’s Theorem Powers Nth roots Summary DE MOIVRE’S THEOREM: POWERS AND ROOTS ALGEBRA 8 INU0114/514 (MATHS 1) Dr Adrian Jannetta MIMA CMath FRAS De Moivre’s Theorem: powers and roots 1 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Objectives This presentation will cover the following: • Recap of polar and exponential form • Understand more clearly the argument of a complex number • De Moivre’s theorem • Using De Moivre’s theorem to find powers and roots De Moivre’s Theorem: powers and roots 2 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Recap: polar and Cartesian form Argand diagrams allow a complex number z to be represented geometrically. The polar form is defined two numbers. yi z = x + yi The modulus (|z| or r) is the distance r of the point from the origin. r The argument (arg z or θ ) is an θ angle measured from the positive x-axis. We can convert from polar to Cartesian form using x = r cos θ y x x and y = r sin θ And from Cartesian form to polar form using y p r = x2 + y 2 and θ = tan−1 x where the argument θ must be in the correct quadrant. De Moivre’s Theorem: powers and roots 3 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary A nice pattern... Consider a complex number z in polar form: z = cos θ + i sinθ 2 Let’s calculate z by squaring both sides: z2 = (cos θ + i sinθ )2 Expand the brackets on the RHS: z2 = cos2 θ + i2 sin2 θ + 2i sinθ cos θ = cos2 θ − sin2 θ + 2i sinθ cos θ We know from trigonometry that cos2 θ − sin2 θ ≡ cos 2θ and 2 sin θ cos θ ≡ sin 2θ . Therefore: z2 = cos 2θ + i sin2θ To summarise: (cos θ + i sinθ )2 ≡ cos 2θ + i sin2θ We can take the power and multiply it with the angle. But is this true for any power/exponent? De Moivre’s Theorem: powers and roots 4 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary De Moivre’s Theorem De Moivre’s theorem is a relationship between complex numbers and trigonometry. Here it is: (cos θ + i sin θ )n = cos nθ + i sin nθ By expanding the LHS and comparing real and imaginary coefficients it is possible to derive trig identities for powers of sine and cosine in terms of compound angles — very useful in calculus. Abraham de Moivre (1667 — 1754) De Moivre’s Theorem: powers and roots De Moivre’s theorem also makes it possible to quickly evaluate powers and nth roots of complex numbers. 5 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Evaluating powers Calculating powers Given the complex number p z = 1 + 3i Use De Moivre’s theorem to evaluate z5 . The modulus is |z| = 2 and arg z = π3 . The polar form of z is z = 2(cos π3 + i sin π3 ) Therefore: z5 = = 5 2(cos π3 + i sin π3 ) 32(cos π3 + i sin π3 )5 By De Moivre’s theorem: 5π z5 = 32(cos 5π 3 + i sin 3 ) Simplify to get the Cartesian form: z5 = = De Moivre’s Theorem: powers and roots p 3 2 i) p 16 − 16 3i 32( 12 − 6 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary The argument revisited yi yi yi z = x + yi r θ z = x + yi r r θ + 2π x z = x + yi θ + 2nπ x x The argument of a complex number arg z = θ is the angle made by the line to positive x-axis (first picture). The angle θ can be increased by 2π and it will still describe the same line in the xy-plane (second picture). Any integer multiple of 2π can be added (or subtracted) to θ without changing the Cartesian form of the complex number. These statements are simply a consequence of the periodic nature of sine and cosine: cos(θ + 2nπ) ≡ cos θ De Moivre’s Theorem: powers and roots and sin(θ + 2nπ) ≡ sin θ 7 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Consider the complex number z = 4(cos π4 + i sin π4 ) = 4∠ π4 . Here is the complex number on an Argand diagram yi z = 4(cos π4 + i sin π4 ) 4 π 4 x If we add 2π (or 4π or 6π or...2nπ) to the angle the point on the Argand diagram will remain unchanged. In this case z is also equivalent to z = 4∠ 9π 4 The angle π 4 , z = 4∠ 17π 4 , z = 4∠ 25π 4 is the principal value. But adding multiples of 2π will give an equivalent angle. De Moivre’s Theorem: powers and roots 8 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary The general polar form of a complex number is z = r(cos(θ + 2πn) + i sin(θ + 2πn)) = r∠(θ + 2πn) and the general exponential form is z = ei(θ +2πn) We have been restricting the allowable values to 0 ≤ θ < 2π. But as we saw earlier, other values of θ are also valid. Multivalued functions? Previously we saw how to calculate logarithms of complex numbers. For example ln(−10) = ln 10 + πi Recall that the definition of a function states that one input (in the domain) should give one output value (in the range). We usually don’t allow other values — like ln10 + 3πi — in order to satisfy that definition. We do however, need to consider those other values of θ to obtain nth roots using De Moivre’s theorem. De Moivre’s Theorem: powers and roots 9 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Calculation of nth roots Definition A number w is a nth root of a number z if wn = z. A few statements to clarify this definition... • The number 2 is the 4th root of 16 because 24 = 16. • The number −2 is also the 4th root of 16 because (−2)4 = 16. • The number 3 is the cube root of 27 because 33 = 27. From the previous example: p p • The number 1 + 3i is the 5th root of 16 − 16 3i The fundamental theorem of algebra tells to expect n roots — not just one. De Moivre’s theorem in this case is 1 θ + 2πn 1 θ + 2πn z n = r n cos + i sin n n where n is an integer. De Moivre’s Theorem: powers and roots 10 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Cube roots of a complex number Find all of the cube roots of 27i. This means we are looking for complex numbers z with the property z = 27i First, we write z in polar form z = 27(cos π2 + i sin π2 ) Take cube roots and apply De Moivre’s theorem: 1 z3 1 27(cos π2 + i sin π2 ) 3 = 1 = 3(cos π2 + i sin π2 ) 3 1 z3 = 3(cos π6 + i sin π6 ) 1 In Cartesian form this is z 3 = p 3 3 2 + 32 i. However, there are other values of z which give z = 27i. De Moivre’s Theorem: powers and roots 11 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary We wrote 27i in polar form as 27i = 27(cos π2 + i sin π2 ) but we could also add 2π to the angle and it would still be 27i: 5π 27i = 27(cos 5π 2 + i sin 2 ) Therefore we have 5π 27i = z = 27(cos 5π 2 + i sin 2 ) If we apply De Moivre’s theorem to this: 1 z3 1 z3 = 1 5π 3 27(cos 5π + i sin ) 2 2 5π = 3(cos 5π 6 + i sin 6 ) 1 p In Cartesian form this is z 3 = − 3 2 3 + 23 i. We found another cube root — different to the first! De Moivre’s Theorem: powers and roots 12 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Add another 2π to the polar form for 27i and we get z 1 z3 1 ∴z3 = 9π 27(cos 9π 2 + i sin 2 ) = 9π 3 27(cos 9π 2 + i sin 2 ) = 3π 3(cos 3π 2 + i sin 2 ) 1 1 In Cartesian form this is z 3 = −3i. We’ve found three cube roots now. Are there any more? Adding another 2π gives us z 1 ∴ z3 1 z3 = 13π 27(cos 13π 2 + i sin 2 ) = 13π 3(cos 13π 6 + i sin 6 ) = p 3 3 2 + 23 i This is a repeat of the first root. If we add more multiples of 2π we will also duplicate the other roots. De Moivre’s Theorem: powers and roots 13 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary The number 27i has three cube roots. They are z1 = p 3 3 2 + 23 i p z2 = − 3 2 3 + 23 i z3 = −3i Or in polar form: z1 = 3∠ π6 , z2 = 3∠ 5π 6 , z3 = 3∠ 3π 2 On an Argand diagram we can see an obvious pattern with these roots: yi z2 z1 x z3 The roots are equally spaced around the circle. The angle between the roots on the diagram is circle is |z| = 3. De Moivre’s Theorem: powers and roots 2π 3 . 14 / 17 The radius of the Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary De Moivre’s Theorem and nth roots. Calculate all of the 4th roots of −64 1 We are trying to find z 4 where z = −64. There will be four roots. In polar form we write −64 = 64(cos π + i sin π) = 64∠π. Therefore the four equations we must solve are: z = 64∠π z = 64∠3π z = 64∠5π z = 64∠7π Applying De Moivre’s theorem to these gives the four roots: p 1 1 z 4 = [64∠π] 4 = 8∠ π4 = 2 + 2i p 1 1 z 4 = [64∠3π] 4 = 8∠ 3π 4 = −2 + 2i p 5π 1 1 z 4 = [64∠5π] 4 = 8∠ 4 = −2 − 2i p 1 1 z 4 = [64∠7π] 4 = 8∠ 7π 4 = 2 − 2i De Moivre’s Theorem: powers and roots 15 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary The Argand diagram of the 4th roots of −64 looks like this: yi z2 z1 x z3 z4 The roots are equally spaced around the circle; the angle between p π them is 2π 4 = 2 . The radius of the circle is |z| = 8. De Moivre’s Theorem: powers and roots 16 / 17 Adrian Jannetta Introduction De Moivre’s Theorem Powers Nth roots Summary Summary Definition A number w is a nth root of a number z if wn = z. In general, to find the nth root of a complex number: • There will be n roots • The n roots will be equally spaced by an angle 2π n . • The polar form will be 1 1 θ + 2πn θ + 2πn z n = r n cos + i sin n n where r is and θ is the principal value (of the first root). De Moivre’s Theorem: powers and roots 17 / 17 Adrian Jannetta