Multiply & Divide Complex Numbers in Trigonometric (Polar) Form
Multiply & Divide Complex Numbers in Trigonometric (Polar) Form - How it Works - Video
Formulas
Formulas:
These are the formulas that we will use to transform a complex number into standard form from trigonometric from or vice versa.
Example 1 - Modulus & Argument 1st Term
Example 1 - Modulus & Argument 1st Term:
Here we have our number in standard form, z = a + b * i. We must convert it to the form z = r * (cosθ + i * sinθ). We know r = sqrt( a2 + b2) and tanθ = b/a.
If we use the a and b of our complex number to form a triangle. The modulus of z or r or the distance between the origin and our point is the hypotenuse. Now we can use a and b as our legs to find our missing side length.
[2 * sqrt(3)]2 + 22 = c2
sqrt([2 * sqrt(3)]2 + 22) = c
sqrt(4 * 3 + 4) = c
sqrt(12 + 4) = c
sqrt(16) = c
c = 4
Pythagorean Theorem
Substituted for a and b
Squared both numbers
Multiplied
Added
Found square root
tanθ = b/a
tanθ = -2 / [2sqrt(3)]
tanθ = -1 / sqrt(3)
tanθ = -sqrt(3) / 3
θ = tan-1(-sqrt(3) / 3)
θ = -π/6
Substituted [2sqrt(3)] for a and -2 for b
Divided
Multiplied top and bottom by sqrt(3)
Used the inverse tangent on both sides
Calculated
We could add 2π to our θ to make it positive, but since value of -π/6 is smaller than 2π so it might be easier to deal with. Remember when we add 2π, we are adding a going a circle one whole turn so we are back to where we started.
We have the theta that we want. We also have our r, so now we can use z = r * (cosθ + i * sinθ). Our final answer is z = 4 * [cos(-π/6) + i * sin(-π/6)].
Example 1 - Modulus & Argument 2nd Term
Example 1 - Modulus & Argument 2nd Term:
Here we have our number in standard form, z = a + b * i. We must convert it to the form z = r * (cosθ + i * sinθ). We know r = sqrt( a2 + b2) and tanθ = b/a.
If we use the a and b of our complex number to form a triangle. The modulus of z or r or the distance between the origin and our point is the hypotenuse. Now we can use a and b as our legs to find our missing side length.
(-1)2 + [sqrt(3)]2 = c2
sqrt{(-1)2 + [sqrt(3)]2} = c
sqrt{1 + 3} = c
sqrt{4} = c
c = 2
Pythagorean Theorem
Substituted for a and b
Squared both numbers
Added
Found square root
tanθ = b/a
tanθ = sqrt(3) / (-1))
θ = tan-1(-sqrt(3) / 1)
θ = 2π/3
Substituted [2sqrt(3)] for a and -2 for b
Used the inverse tangent on both sides
Calculated
We have the theta that we want. We also have our r, so now we can use z = r * (cosθ + i * sinθ). Our final answer is z = 2 * [cos(2π/3) + i * sin(2π/3)].
Example 1 - Product
Example 1 - Product:
When we multiply different complex numbers we multiply the moduli and add the arguments. Now we can multiply z1 = 4 * cos(-π/6) + i * sin(-π/6)] and z2 = 2 * cos(2π/3) + i * sin(2π/3)].
z1 * z2 = 4 * 2 * [cos(-π/6 + 2π/3) + i * sin(-π/6 + 2π/3)]
z1 * z2 = 8 * [cos(-π/6 + 4π/6) + i * sin(-π/6 + 4π/6)]
z1 * z2 = 8 * [cos(3π/6) + i * sin(3π/6)]
z1 * z2 = 8 * [cos(π/2) + i * sin(π/2)]
z1 * z2 = 8 * [0 + i * 1]
z1 * z2 = 8i
Multiplied the moduli and added the arguments
Multiplied the moduli and found the common denominator
Added the arguments
Simplified the arguments
Calculated
Distributed
Example 1 - Quotient
Example 1 - Quotient:
When we divide different complex numbers we divide the moduli and subtract the arguments. Now we can divide z1 = 4 * cos(-π/6) + i * sin(-π/6)] and z2 = 2 * cos(2π/3) + i * sin(2π/3)].
z1 / z2 = 4 / 2 * [cos(-π/6 - 2π/3) + i * sin(-π/6 - 2π/3)]
z1 / z2 = 2 * [cos(-π/6 - 4π/6) + i * sin(-π/6 - 4π/6)]
z1 / z2 = 2 * [cos(-5π/6) + i * sin(-5π/6)]
z1 / z2 = 2 * [-sqrt(3)/2 + i * (-1/2)]
z1 / z2 = 2 * [ {-sqrt(3) - i} / 2} ]
z1 / z2 = -sqrt(3) - i
Divided the moduli and subtracted the arguments
Divided the moduli and found the common denominator
Subtracted the arguments
Calculated
Combined fractions
Distributed