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8.3: Polar Form of Complex Numbers Real numbers can be graphed on a number line. A complex number, a + ib, has two parts, the real part, a, and the imaginary part, b. It can not be shown on a number line, but it can be represented by a point in the complex plane, the plane where the horizontal axis is the real axis (and is simply the real number line) and the vertical axis is the imaginary axis (and is labeled with pure imaginary numbers, i, 2i, etc.). Graph these numbers in the complex plane: A. 3 + 2i B. 5 – 3i C. -2 + 4i D. 3 – 2i E. 5 + 3i F. -2 – 4i |z| is called the modulus of z, or the absolute value of z. |z| = Sketch the following sets in the complex plane: 22. {z = a + bi| a > 1, b > 1} 26. {z| 2 ≤ |z| ≤ 5 } a2 + b 2 Since we can represent any complex number, z = a + ib, as the point (a,b) in the complex plane, it follows that we can also represent any complex number in the complex plane as a point using polar coordinates. Similar to our work in Chapter 7, we find that the real part of z can be represented by a = r cosθ, and the imaginary part of z as b = r sinθ , where r = |z| and tanθ = b/a. Then z = r (cosθ + i sinθ) The number r is called the modulus of z and θ is called the argument of z. [Note that r is always non-negative.] Write the following complex numbers in polar form with the argument (θ) between 0 and 2π. 30. 1+i 3 34. -1 + i 38. 4 46. 2(1 – i) 48. -3 – 3i 52. -π i When adding or subtracting complex numbers, the rectangular form (a + ib) is generally the easier one to use. However, when multiplying, dividing, raising to powers, or finding roots of complex numbers, the polar form is much nicer. Consider what happens when we multiply two complex numbers. Let z = r(cosα + i sinα) zw and let w = ρ(cosβ + i sinβ) = r(cosα + i sinα) ρ(cosβ + i sinβ) = r ρ (cosα + i sinα) (cosβ + i sinβ) = r ρ [cosα cosβ + cosα (i sinβ) + (i sinα) cosβ + (i sinα)(i sinβ)] = r ρ [cosα cosβ + i(cosα sinβ + sinα cosβ) + i2 sinα sinβ ] = r ρ [cosα cosβ – sinα sinβ + i (cosα sinβ + sinα cosβ)] = r ρ [cos(α + β) + i sin(α + β)] To multiply 2 complex numbers in polar form, we multiply their moduli and add their arguments. For example: [2(cos23° + i sin23°)] [7(cos118° + i sin118°)] = 14(cos141° + i sin141°) Analogously, to divide two complex numbers, one divides their moduli and subtracts their arguments: z = r(cosα + i sinα) and w = ρ(cosβ + i sinβ), then z r = w ρ [cos(α - β) + i sin(α - β)] z1 #54 & 58: Find the product z1 z2 and the quotient z . Express your answer in polar form. 2 #54: z1 = π cos 4 +i π sin 4 z2 = 3π cos 4 +i 3π sin 4 #58. z1 = 2 (cos75° + i sin75°) z2 = 3 2 (cos60° + i sin60°) z1 #62 & #66: Write z1 and z2 in polar form, and then find the product z1 z2 and the quotients z 2 1 and z 1 #62: z1 = 2 - i 2 z2 = 1 – i #66: z1 = 4 3 - 4 i z2 = 8 i Raising a number to a positive integer power is simply successive multiplication. This gives rise to De Moivre’s Theorem: If z = r(cos θ + i sin θ), then for any integer n, zn = rn(cos nθ + i sin nθ) To take the nth power of a complex number, we take the nth power of the modulus and multiply the argument by n. #76 & # 70: Find the indicated power using De Moivre’s Theorem 15 -1 3 #76: 2 - i 2 -1 3 4π 4π i 2 = cos 3 + i sin 3 2 4π 115 = 1 and 3 15 = 20π 4π The modulus is 1 and the argument is 3 . 15 -1 3 i 2 2 = 1 (cos20π + i sin20π) = cos0 + i sin0 = 1 (since 0 is coterminal with 20π) #70: (1 – i 3 )5 nth Roots of Complex Numbers De Moivre’s Theorem also gives us a method to determine the nth roots of any complex number. If z = r(cosθ + i sinθ ) and n is a positive integer, then z has the n distinct nth roots: θ + 2kπ θ + 2kπ + i sin n n wk = r1/n cos for k = 0, 1, 2, . . . , n-1. Note: 1/n 1. The modulus of each nth root is r . 2. The argument of the first root is θ/n. 3. We repeatedly add 2π/n to get the argument of each successive root. These observations show that, when graphed, the nth roots of z are spaced evenly on the circle of radius r 1/n . Examples: Find the indicated roots and graph the roots in the complex plane. #82. The cube roots of 4 3 + 4i. z = 4 3 + 4i , then |z| = r = 8 and θ = 60° (We’ll use degrees to make it easy on ourselves.) 60° + k360° 1/3 60° + k360° wk = 8 cos + i sin for k = 0, 1, 2 3 3 w0 = 2[cos20° + i sin20°] w1 = 2[cos140° + i sin140°] w2 = 2[cos260° + i sin260°] These graph as 3 points on the circle r = 2, one at θ = 20°, one at θ = 140°, and one at θ = 260°. #84: The fifth roots of 32. 1/5 32 = 32(cos0° + i sin0°), so r = 32, and θ = 0°. r =2 and θ/5 = 0° 360°/5 = 72°, so we are all set. w0 = 2[cos0° + i sin0°] = 2 w1 = 2[cos72° + i sin72°] w2 = 2[cos144° + i sin144°] w3 = 2[cos216° + i sin216°] w4 = 2[cos288° + i sin288°] These graph as: #86: Find the cube roots of 1 + i. and graph them. 3 Solve this equation: z – 4 3 – 4i = 0