Complex Numbers

Written by tutor Colin D.

How to Envision Complex Numbers Graphically: The Complex Plane

How to Envision Complex Numbers on a Graph

  • The complex number x + yi corresponds to the point with coordinates (x, y)
  • The x-axis is the real axis
  • The y-axis is the imaginary axis
  • Real numbers are associated with points on the x-axis
    For example: x = x + 0i <- -=””> (x,0)
  • Imaginary numbers are associated with points on the y-axis
    For example: yi = 0 + yi <- -=””> (0,y)

How to Find a Point (P) in the Complex Plane

  • Any point in the complex plane can be identified by the coordinate pair (r, θ)
  • r = distance from the origin to point P (i.e., line segment OP)
  • θ = angle from the positive x-axis (between Quadrants I and IV) to segment OP
  • All points on the terminal side can be expressed as (r cos θ, r sin θ)
    -Because cos θ = adjacent/hypotenuse, and hypotenuse = r,
    to solve for θ, one would proceed: cos θ = x/r.
    Solving this for x would result in x = r cos θ
    -Likewise, because sin θ = opposite/hypotenuse, solving for θ
    would result in sin θ = y/r.

    Solving this for y would result in y = r sin θ

  • Piecing it altogether:
    -If we have complex number x + yi
    -Then P has coordinates (x,y)
    -And x = r cos θ, y = r sin θ

Trigonometric (Polar) Form

Trigonometric (Polar) Form on a Graph

  • The trigonometric form of “x + yi” is r(cos θ + i sin θ)
    -This can be derived from earlier equivalences. Because when we had
    x + yi, we found x = r cos θ and
    y = r sin θ, we can replace x and y with r cos θ
    and r sin θ, respectively:
    x + yi = (r cos θ) + (r sin θ)i
    -By factoring out the “r” and multiplying by the “i,” this turns into:
    r(cos θ + i sin θ)
  • r = Modulus or Absolute Value
    r = (x2 + y2)1/2
    r = must be NON-negative
  • θ = Argument of the complex number
    -Any angle coterminal with θ is also an argument for the same complex number
    tan θ = y/x -> θ = arc tan (y/x)

Rectangular (Standard) Form

  • Rectangular form is “x + yi”

How to Change from Rectangular Form to Trigonometric Form

Change from Rectangular to Trigonometric Form on a Graph

  • If A = 2 + 2i
    -First, find “r.” Remember, “r” – called the modulus – is the absolute value of the
    hypotenuse formed by
    sides “x” and “y”
    r = √(22 + 22)
    r = √(4+4)
    r = √8
    r = 2√2
    -Next, find θ. Remember, θ is called the argument, and is found through the following equation:
    tan θ = y/x because the tangent of the angle formed by “r” and the “x-axis” equals the opposite side
    divided by the adjacent side (i.e., the y-value divided by the x-value).

    θ = arc tan (y/x)
    θ = arc tan (2/2)
    θ = arc tan (1)

    θ = 45°

  • Then trigonometric form is found by plugging in “r” and “θ”:
    A = r(cos θ + i sin θ)
    A = 2√2(cos 45° + i sin 45°)

How to Change from Trigonometric Form to Rectangular Form

Change from Trigonometric to Rectangular Form on a Graph

  • If B = 3√3 (cos 330° + i sin 330°)
    r = 3√3
    cos 330° = √(3/2)
    sin 330° = -1/2
  • Then 3√3 (√(3/2) + -1/2 i) -> 9/2 – i(3√3)/2

How to Express Complex Numbers in Proper Trigonometric Form

Proper Trigonometric Form on a Graph

  • Always remember a few essentials about proper trigonometric form:
    -The modulus (r) must always be non-negative
    It is the absolute value of the diagonal from the point itself to the origin.
    -The parenthetical expression must be of the form: cos θ + i sin θ.
    Make sure each term is written as a positive amount.
  • Example: z = 2(cos 30° – i sin 30°)
    -First, express z in rectangular form:
    2(√(3/2) – 1/2 i) -> √3 – 1i
    -Thus, on a graph, this would consist of moving √3 units to the right, 1 unit down, resulting in a point in
    Quadrant IV.
    r = √((√3)2 + 12) -> √4 -> √2

    Using tan θ = y/x, we derive:

    tan θ = 1/√3 -> arctan 1/√3 = -30° -> hence, θ = -30°
    -Finally, substitute:

    z = 2[cos (-30°) + i sin (-30°)]

Multiplication & Division in Trigonometric Form

  • NOTE: While rectangular form makes addition/subtraction of complex numbers easier to conceive of, trigonometric form is the best method
    of conceiving of complex for multiplication/division purposes.
  • If you intend to multiply two complex numbers, z1 = r1 (cos θ1 + i sin θ1), and z2 = r2 (cos θ2 + i sin θ2), the product
    is derivable by following a few simple steps:
    Multiply the moduli to find the product modulus: r1 times r2
    Add the arguments to find the sum argument: cos (θ1 + θ2)
    + i sin (θ1 + θ2)
    -Multiply the product modulus by the sum argument: r1r2 [cos (θ1 + θ2) +
    i sin (θ1 + θ2)]
  • To divide two complex numbers:
    Divide the moduli to get the quotient modulus: r1/r2
    Subtract the arguments to get the difference argument: cos (θ1 – θ2)
    + i sin (θ1 – θ2)
    -Multiply the quotient modulus by the difference argument: r1/r2 [cos (θ1 – θ2)
    + i sin (θ1 – θ2)]
  • Example:
    z1 = √(3/2 + (1/2)i
    z2 = -2 – 2i
    Find z1 * z2:
    (1) Express each in trigonometric form
    z1 = 2(cos 30° + i sin 30°)
    z2 = 2√2(cos 225° + i sin 225°)
    (2) Multiply moduli:
    2 * 2√2 = 4√2
    (3) Add arguments:
    cos(30° + 225°) + i (sin 30° + 225°)
    (4) Triangular Form = 4√2 [cos(30° + 225°) + i (sin 30° + 225°)]
    4√2[cos (255°) + i (sin 255°)]
    To find in Rectangular Form, evaluate the cos 255° and sin 255° and simplify:
    4√2 [cos (255°) + i (sin 255°)]
    With the sum and difference formulae:
    cos (a+b) = cos a cos b – sin a sin b
    sin (a+b) = sin a cos b + sin b cos a
    With calculator:
    cos 255° = -.2588
    sin 255° = -.9659
    -1.464 – 5.464i

DeMoiver’s Theorem

  • By repeating the multiplication procedure outlined just above, one may derive DeMoivre’s Theorem, which allows us to compute powers
    and roots of complex numbers.
  • To illustrate, if we were to continue to multiply z = r (cos θ + i sin θ) by itself, we’d get:
    z2 = r2 (cos 2θ + i sin 2θ)
    z3 = r3 (cos 3θ + i sin 3θ)
    z4 = r4 (cos 4θ + i sin 4θ)
  • For negative exponents, it unfolds in the following pattern:
    z-1 = r-1 [(cos(-θ)) + i sin (-θ)]
    z-2 = r-2 [(cos(-2θ)) + i sin (-2θ)]
  • Formally stated as a rule, DeMoivre’s Theorem reads:
    zn = rn (cos nθ + i sin nθ)
  • EXAMPLE:
    (1 + √3i)5
    -In trigonometric form:
    2(cos 60° + i sin 60°)
    -Apply DeMoivre’s Theorem:
    25 [cos 5(60°) + i sin 5(60°)]
    32 (cos 300° + i sin 300°)
    32 (1/2 + i(-√3)/2)
    16 – (16√3)i

Roots of Complex Numbers

Calculating roots of complex numbers

  • Some basics about visualizing the roots of complex numbers:
    -The n roots of a complex number all lie on the circle formed within the complex plane with center at the origin and radius = (r)(1/n)
    -The n roots on said circle are all equally spaced, beginning at K = 0 and proceeding until k = n-1, progressing at arguments (i.e., intervals) differing by 360°/n
  • Formula:
    -Given any positive integer, n, then the nonzero complex number z (where z = r (cos θ + i sin θ)) has exactly n distinct nth roots, given by the following equation, in which k = 0, 1, 2,…., (n-1):
    W (sub k) = (r)^n [cos (θ/n + k * 360°/n) + i sin (θ/n + k * 360°/n)]
  • Example: Find the 6th roots of 5 + 12i
    (1) Write 5 + 12i in trigonometric form:

    r = √(52 + 122) = 13
    θ = arctan (12/5) ~ 67.38°
    Thus, 5 + 12i = 13(cos θ + i sin θ)
    (2) Since we’re looking for sixth roots (n = 6), we replace n with 6 and simplify:
    W (sub k) = 131/6 [cos(θ/6 + k * 360/6) + i sin (θ/6 + k * 360/6)]
    W (sub k) = 1.533 [cos (67.38/6 + k * 60) + i sin 11.23 + 60k)]
    W (sub k) = 1.533 [cos(11.23 + 60k) + i sin (11.23 + 60k)]
    (3) Plug in the various values of k up to (n-1) where n = 6 (because of the sixth roots).
    K = 0 -> 1.504 + .299i
    K = 1 -> .493 + 1.451i
    K = 2 -> -1.01 + 1.153i
    K = 3 -> -1.504 + -.299i
    K = 4 -> -.493 + -1.451i
    K = 5 -> 1.01 + -1.153i

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