Complex Numbers and Quadratic Equations

Roots of the Complex Numbers and Geometric Interpretations


Tips:

  • If z = r (cos θ + i sin θ) and ‘n’ is a positive integer, then \tt z^{\frac{1}{n}} = r^{\frac{1}{n}}\left[\cos\left(\frac{2k\pi + \theta}{n}\right) + i \sin \left(\frac{2k\pi + \theta}{n}\right)\right] where K = 0, 1, 2, 3, ……. (n − 1)
  • 1, ω, ω2 are the cube roots of unity
  • 1 + ω + ω2 = 0 ; ω3 = 1, ω3n = 1 ; ω3n + 1 = ω ; ω3n + 2 = ω2
  • If a + bω + cω2 = 0, then a = b = c, provided a, b, c are real
  • ±1, ±i are the fourth roots of unity

Tricks:

  • Square root of z = a + ib are \tt \pm \left[\sqrt{\frac{|z| + a}{2}} + i \sqrt{\frac{|z| - a}{2}}\right] for b > 0 \tt \pm \left[\sqrt{\frac{|z| + a}{2}} - i \sqrt{\frac{|z| - a}{2}}\right]for b < 0
  • \tt \sqrt{z} + \sqrt{\overline{z}} = \pm \sqrt{2}\left\{\sqrt{|z| + a}\right\} where imaginary (z) > 0
  • \tt \log i = \log {e}^{\frac{i \pi}{2}} = \frac{i \pi}{2}
  • |z – z1| = |z – z2| ⇒ Locus is perpendicular bisector of the line segment joining z1 and z2 [z1, z2 are fixed points, z is variable]
  • |z – z1| + |z – z2| = constant ⇒ Locus of ‘z’ is an ellipse [z1, z2 are fixed points, z is variable]
  • |z – z1| + |z – z2| = |z1 – z2| ⇒ line segment [z1, z2 are fixed points, z is variable]
  • |z – z1| – |z – z2| = Constant ⇒ Hyperbola [z1, z2 are fixed points, z is variable]
  • |z – z1| = k |z – z2| (K ≠ 1) ⇒ circle [z1, z2 are fixed points, z is variable]
  • \tt arg \left(\frac{z - z_{1}}{z - z_{2}}\right) = \pm \frac{\pi}{2}\Rightarrow Locus of ‘z’ is a circle with z1 and z2 as the vertices of diameter.
  • \tt arg \left(\frac{z - z_{1}}{z - z_{2}}\right) = o (or) \pi \RightarrowLocus of ‘z’ is a straight line passing through z1 and z2
  • In general ||z1| − |z2|| ≤ |z1 + z2| ≤ |z1| + |z2|
  • If \tt |z + \frac{1}{z}| = a, the greatest and least values of |z| are respectively \tt \frac{a+ \sqrt{a^{2} + 4}}{2} \ and \ \frac{-a + \sqrt{a^{2} + 4}}{2}
  • The equation |z – z1|2 + |z – z2|2 = k [k is real number] will represent a circle with centre at \tt \frac{1}{2}(z_{1} + z_{2}) and radius \tt \frac{1}{2}\sqrt{2k - \left(z_{1} - z_{2}\right)^{2}} provided \tt k \geq \frac{1}{2}|z_{1} - z_{2}|^{2}
  • If z1, z2, z3 …….zn have the same amplititude then |z1| + |z2| + ……… |zn| = |z1 + z2 + z3 + …. + zn|
  • If ‘3’ points z1, z2, z3 are connected by relation az1 + bz2 + cz3 = 0 where a + b + c = 0, then the ‘3’ points are collinear.
  • If ‘3’ complex number are in A.P. then they lie on a straight line in the complex plane.
  • If ‘z’ is a complex number, then ez is periodic

Part1: View the Topic in this video From 20:52 To 51:26

 

Part2: View the Topic in this video From 00:40 To 52:08

Part3: View the Topic in this video From 00:40 To 52:20

Part4: View the Topic in this video From 00:40 To 55:00

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  • The area of the triangle whose vertices are z, iz and z + iz is \tt \frac{1}{2} |z|^{2}
  • The area of the triangle with vertices z, ωz and z + ωz is \tt \frac{\sqrt{3}}{4} |z|^{2}
  • If z1, z2, z3 be the vertices of an equilateral triangle and zo be the circumcentre, then \tt z_{1}^{2} + z_{2}^{2} + z_{3}^{2} = 3z_{0}^{2}
  • If z1, z2, z3 be the vertices of a triangle, then the triangle is equilateral iff (z1 − z2)2 + (z2 − z3)2 + (z3 − z1)2 = 0 (or) \tt z_{1}^{2} + z_{2}^{2} + z_{3}^{3} = z_{1}\cdot z_{2} + z_{2}\cdot z_{3} + z_{3} \cdot z_{1} (or) \frac{1}{z_{1} - z_{2}} + \frac{1}{z_{2} - z_{3}} + \frac{1}{z_{3} - z_{1}} = 0
  • x2 + y2 = (x + iy)(x − iy)
  • x3 + y3 = (x + y)(x + yω)(x + yω2)
  • x2 + xy + y2 = (x − yω)(x – yω2) in particular x2 – x + 1 = (x − ω)(x –ω2)
  • x2 + y2 + z2 – xy – yz – zx = (x + yω + zω2) (x + yω2 + zω)
  • x3 + y3 + z3 – 3xyz = (x + y + z) (x + ωy + ω2z) (x + ω2y + ωz)
  • Equation of the perpendicular bisector is \tt z(\overline{z_{1}} - \overline{z_{2}}) + \overline{z}(z_{1} - z_{2}) = |z_{1}|^{2} - |z_{2}|^{2}
  • Equation of a straight line in parametric form z = tz1 + (1 − t) z2 where ‘t’ ∈ R
  • Non-parametric form is \tt \begin{bmatrix}z & \overline{z} & \overline{1}\\z_{1} & \overline{z_{1}} & 1 \\z_{2} & \overline{z_{2}} & 1 \end{bmatrix} = 0