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EXERCISES - COMPLEX NUMBERS


Write in algebraic form (1+i)/(1-i); z⋅z = |z|2 = a2+b2; (1+i)/(1-i)⋅(1+i)/(1+i) = (1+i)2/(12+12) = (1-1+2i)/(1+1) = 2i/2 = i

z2+z = 0; z = a+ib; (a+ib)2+(a-ib) = 0 ⇔ a2-b2+i2ab+a-ib = 0 ⇔ (a2-b2+a)+i(2ab-b) = 0 ⇔ a2-b2+a = 0, 2ab-b = 0; 2ab-b = 0 ⇔ b(2a-1) = 0 ⇔ b = 0, a = 1/2; b = 0, a2 + a = 0 ⇔ a(a+1) = 0 ⇔ a = 0, a = -1; z1 = 0, z2 = -1; a = 1/2, 1/4 - b2 + 1/2 = 0 ⇔ -b2 + 3/4 = 0 ⇔ -b2 = - 3/4 ⇔ b2 = 3/4 ⇔ b = ±√(3/4) = ±√(3)/2; z3 = 1/2+i√(3)/2, z4 = 1/2-i√(3)/2

(z)4 = |z|; z = ρ(cos(θ)+i⋅sin(θ)); |z| = ρ; z = ρ(cos(θ)-i⋅sin(θ)); Considering De Moivre's formula, ρ4(cos(4θ)-i⋅sin(4θ)) = ρ(cos(0)-i⋅sin(0)); ρ4 = ρ; 4θ = 0+2kπ k ∈ ℤ; ρ4 - ρ = 0 ⇔ ρ(ρ3-1) = 0 ⇔ ρ = 0, ρ = 1; θ = kπ/2 k ∈ ℤ; ρ = 0, z1 = 0; ρ = 1 and k = 0, z2 = 1(cos(0)+i⋅sin(0)) = 1; ρ = 1 and k = 1, z3 = 1(cos(π/2)+i⋅sin(π/2)) = i; ρ = 1 and k = 2, z4 = 1(cos(π)+i⋅sin(π)) = -1; ρ = 1 and k = 3, z5 = 1(cos(3π/2)+i⋅sin(3π/2)) = -i

3√(i-1); z = -1+i; |z| = √(a2+b2) = √(-12+12) = √2; arg(z) = 3π/4; zk = ρ1/n(cos((θ+2kπ)/n)+i⋅sin((θ+2kπ)/n), k = 0, 1, ..., n-1; zk = (21/2)1/3(cos(((3/4)π+2kπ)/3)+i⋅sin(((3/4)π+2kπ)/3)); k = 0, z0 = 21/6(cos(π/4)+i⋅sin(π/4)) = 21/6(√2/2+i⋅√2/2); k = 1, z1 = 21/6(cos(((3/4)π+2π)/3)+i⋅sin(((3/4)π+2π)/3)) = 21/6(cos((11/12)π)+i⋅sin((11/12)π)); k = 2, z2 = 21/6(cos(((3/4)π+4π)/3)+i⋅sin(((3/4)π+4π)/3)) = 21/6(cos((19/12)π)+i⋅sin((19/12)π))

z3-z|z|2+z = 0; z(z2-|z|2+1) = 0; z = 0; z2-|z|2+1 = 0; z = a+ib a,b ∈ ℝ; |z| = √(a2+b2); (a+ib)2-(a2+b2)+1 = 0 ⇔ a2+2abi-b2-a2-b2+1 = 0 ⇔ -2b2+1+2abi = 0 ⇔ {-2b2+1 = 0 ⇔ b = ± 1/√2, 2ab = 0 ⇔ {a = 0, b = 0}}; a = 0, b = ± 1/√2, z1 = (1/√2)i, z2 = -(1/√2)i; b = 0, b = ± 1/√2 impossible; z0 = 0, z1 = (1/√2)i, z2 = -(1/√2)i

z5 = 1/(1+i)3; z = ρ(cos(θ)+i⋅sin(θ)); z = ρ(cos(θ)-i⋅sin(θ)); Considering De Moivre's formula, z5 = ρ5(cos(5θ)-i⋅sin(5θ)); 1/(1+i) = (1/(1+i))((1-i)/(1-i)) = (1-i)/(12+12) = (1-i)/2 = (1/2)-(1/2)i; ρ = |1/(1+i)| = √((1/2)2+(1/2)2) = √((1/4)+(1/4)) = √(2/4) = √(1/2) = 1/√2; arg(1/(1+i)) = (7/4)π; 1/(1+i) = 1/√2(cos((7/4)π)+i⋅sin((7/4)π)); (1/(1+i))3 = (1/21/2)3(cos(3⋅(7/4)π)+i⋅sin(3⋅(7/4)π)) = 1/23/2(cos((21/4)π)+i⋅sin((21/4)π)) = 1/23/2(cos((5/4)π)+i⋅sin((5/4)π)), (21/4)π = (16/4)π + (5/4)π = 2(2π)+(5/4)π, φ1 = φ+2kπ, k ∈ ℤ ⇒ cos(φ1) = cos(φ), sin(φ1) = sin(φ); z5 = 1/(1+i)3 ⇔ ρ5(cos(5θ)-i⋅sin(5θ)) = 1/23/2(cos((5/4)π)+i⋅sin((5/4)π)); ρ5(cos(-5θ)+i⋅sin(-5θ)) = 1/23/2(cos((5/4)π)+i⋅sin((5/4)π)); ρ5 = 1/23/2; -5θ = (5/4)π+2kπ, k ∈ ℤ ⇔ θ = -(5/20)π-(2/5)kπ = -(5/20)π+(2/5)kπ, k ∈ ℤ; k = 0, θ0 = -(5/20)π; k = 1, θ1 = -(5/20)π+(2/5)π = (3/20)π; k = 2, θ2 = -(5/20)π+(2/5)2π = -(5/20)π+(4/5)π = -(5/20)π+(16/20)π = (11/20)π; k = 3, θ3 = -(5/20)π+(2/5)3π = -(5/20)π+(6/5)π = -(5/20)π+(24/20)π = (19/20)π; k = 4, θ4 = -(5/20)π+(2/5)4π = -(5/20)π+(8/5)π = -(5/20)π+(32/20)π = (27/20)π; ∃ k | -(5/20)π+2kπ = (27/20)π ?, ∃ k | -(5/20)π+(40/20)kπ = (27/20)π ?, False; k = 5, θ5 = -(5/20)π+(2/5)5π = -(5/20)π+(10/5)π = -(5/20)π+(40/20)π = (35/20)π; ∃ k | -(5/20)π+2kπ = (35/20)π ?, ∃ k | -(5/20)π+(40/20)kπ = (35/20)π ?, true for k = 1; There are 5 different solutions: ρ5 = 1/23/2, k = 0 → θ0 = -(5/20)π, k = 1 → θ1 = (3/20)π, k = 2 → θ2 = (11/20)π, k = 3 → θ3 = (19/20)π, k = 4 → θ4 = (27/20)π