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Calculus Notes: Complex Numbers Jonas Oppenheim† MATH 022 - Calculus II August 8, 2016 1 Standard Form Standard Form: z = a + bi √ where a, b ∈ R and i = −1. If a = 0, then the complex number is pure imaginary. If b = 0, then the complex number is real. Note that R ⊆ C. If z = a + bi and a, b ∈ R, then the conjugate of z is z̄ = a − bi and, as is seen figure 1, the modulus (or absolute value) of z is √ |z| = a2 + b2 1.1 Arithmetic Let w = −5 + 6i z = 3 − 8i. 1.1.1 Addition Simply add the real and imaginary terms independently: w + z = (−5 + 3) + (6i − 8i) = −2 − 2i. † Jonas N. Oppenheim, Calculus II, University of Vermont, Burlington, VT 05405; [email protected]; http://www.jonasoppenheim.com 1 Figure 1: The modulus of z is |z|. 1.1.2 Subtraction Subtract the real and imaginary terms independently: w − z = (−5 − 3) + (6i + 8i) = −8 + 14i. 1.1.3 Multiplication Grit your teeth and distribute: w × z = (−5 + 6i)(3 − 8i) = −15 + 40i + 18i − 48i2 = 33 + 58i. Note that since i = 1.1.4 √ −1, i2 = −1. Division Multiply the numerator and denominator by the conjugate of the denominator: w −5 + 6i = z 3 − 8i (−5 + 6i)(3 + 8i) = (3 − 8i)(3 + 8i) −15 − 40i + 18i + 48i2 = 9 + 24i − 24i − 64i2 −63 − 22i = 73 −63 22i = − 73 73 Note that if z = a + bi, then z × z̄ = (a + bi)(a − bi) = |z|2 . 2 2 Polar Form Polar form: z = r(cos(θ) + i sin(θ)) where r = |z| θ = Arg(z). Alternatively, Euler’s Formula shows that the polar form may also be given as z = reiθ . (a) T1 (x) (b) T2 (x) Figure 2: Comparing 3 Example Problem Let z = 5 − 5i. a) Express z in polar form. Begin by plotting z on the complex coordinate plane. Recall that polar form is z = r(cos(θ) + i sin(θ)) and so we need to find r and θ. In order to find r, we use the Pythagorean 3 theorem. In this case, we find that r 2 = a2 + b 2 r2 = (5)2 + (−5)2 √ r = 50 √ = 5 2. Next, we find θ by recognizing the type √ of triangle being formed on the complex coordinate plane. In this case, we have a 1, 1, 2 triangle and so θ = 45 degrees = π4 . Note, though, or −π . Now that we know r and θ, that because r is positive, θ must be given as either 7π 4 4 √ 7π 7π we know that the polar form of 5 − 5i is 5 2(cos( 4 + i sin( 4 )). b) Find z 12 . We will proceed using the Euler form of z, so that √ 7π z = 5 2ei 4 . Raising both sides to the power of 12, we have √ 7π z 12 = (5 2ei 4 )12 = 512 26 ei21π Now put the result back into polar form: 512 26 ei21π = 512 26 (cos(21π) + i sin(21π)) = 512 26 (−1 + i0) = −(512 26 ). 4 Euler’s Formula ex = ∞ X xn n=0 iθ e = ∞ X (iθ)n n=0 n! =1+ n! iθ i2 θ2 i3 θ3 i4 θ4 i5 θ5 i6 θ6 i7 θ7 + + + + + + + ... 1! 2! 3! 4! 5! 6! 7! iθ θ2 iθ3 θ4 iθ5 θ6 iθ7 − − + + − − + ... 1! 2! 3! 4! 5! 6! 7! θ2 θ4 θ6 θ θ3 θ5 θ7 = (1 − + − + . . . ) + i( − + − + ...) 2! 4! 6! 1! 3! 5! 7! =1+ 4 eiθ = cos(θ) + i sin(θ) Now strap yourself in and consider the case where θ = π. Since cos(π) = −1 and sin(π) = 0, the clouds part, we are bathed in sunlight, and a stone tablet descends from the heavens that reads eiπ = −1. 5