Embedded Files
Multiply & Divide Complex Numbers in Trigonometric (Polar) Form - How it Works - Video
Multiply & Divide Complex Numbers in Trigonometric (Polar) Form - How it Works - Video
Formulas
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
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
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
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
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
Live Worksheet
Live Worksheet
Page updated
Google Sites
Report abuse