complex numbers

• Euler’s relation: e^iθ = cos θ + i sin θ
• Replacing θ with -θ: e^(-iθ) = cos θ - i sin θ
• Subtracting the two relations: e^iθ - e^(-iθ) = 2i sin θ
• Exponential form of complex numbers: z = r*e^(iθ)
• sin θ = 1/2i (e^iθ - e^(-iθ))
• Multiplying and dividing complex numbers: z1*z2 = r1*r2*e^(i(θ1 + θ2))
• De Moivre’s theorem: (r(cos θ + i sin θ))^n = r^n (cos(nθ) + i sin(nθ))
• Trigonometric identities:

  • z^(n+1) + z^(n-1) = 2 cos nθ
  • z^(n+1) - z^(n-1) = 2i sin nθ
• Sums of complex series: S_n = a(1 - r^n)/(1 - r), S_∞ = a/(1 - r)
• To find the n roots of a complex number, you solve the equation z^n = w
• The equation z^n = w has n distinct solutions, where z and w are non-zero complex numbers and n is a positive integer
• De Moivre’s theorem is used to find the roots of a complex number
• An nth root of unity is a solution to the equation z^n = 1
• If you know one root of a complex number with n roots, you can find the other roots by multiplying by an nth root of unity
• The n roots of a complex number lie at the vertices of a regular n-gon with its center at O
• The roots of unity are useful for solving geometric problems
• The point P(√3, 1) lies at one vertex of an equilateral triangle with the center at the origin
• To find the coordinates of the other vertices of the triangle, use the formula z1z2 = |z1||z2| arg(z1) - arg(z2)
• The modulus of a complex number is found by multiplying each modulus
• The argument of a complex number is found by adding the arguments together
• To simplify a complex number division, divide the magnitudes and subtract the arguments
• The modulus argument form is used to write a complex number in the form x + iy
• The n-th root of a complex number is given by z^n = cos(nθ) + i sin(nθ)
• The -n-th root of a complex number is given by z^-n = cos(nθ) - i sin(nθ)
• The sum of z^n and z^-n is 2cos(nθ)
• The exponential form of a complex number is z = √2 e^(iπ/2)
• Raising both sides of z + 1/z = 2cos(θ) to the fifth power results in z^5 + 1/z^5 = 32cos(5θ)
• The binomial expansion is used to expand z + 1/z to the fifth power
• The simplified form of z^5 + 1/z^5 is 2cos(5θ) + 10cos(3θ) + 20cos(θ)
• 32 cos 5 𝜃 = 2 cos(5 𝜃) + 10 cos(3 𝜃) + 20 cos(𝜃)
• �� + ��𝑄 = 𝑒6𝑖(��−13𝑖2 − 𝑒13𝑖2) − 2�� sin(𝜃/2) = 𝑒6𝑖(𝑒13��2 − 𝑒−13��2) / 2𝑖 sin(𝜃/2)
• 𝑧4 = 2 − ��(2√3)
• 𝑧 = 4^(1/4) (cos(−𝜋/3) + 𝑖 sin(−��/3))
• 𝑧 = √2 (cos(−��/12) + 𝑖 sin(−��/12))
• 𝑧 = √2 (cos(−7𝜋/12) + 𝑖 sin(−7𝜋/12))
• �� = 1 + cos(𝜃) + cos(2𝜃) + cos(3𝜃) + ⋯ + cos(12��) + ⋯
• 𝑄 = sin(𝜃) + sin(2𝜃) + sin(3𝜃) + ⋯ + sin(12��) + ⋯
• �� + ��𝑄 = 1 + (cos(𝜃) + 𝑖 sin(𝜃)) + (cos(2𝜃) + 𝑖 sin(2𝜃)) + ⋯
• �� + ��𝑄 = 1(1 − (𝑒𝑖)13) / (1 − 𝑒−𝑖) = 1 − 𝑒13�� / (1 − 𝑒𝑖)
• 𝑧 = √3 + 𝑖