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

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

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 - Multiply

Example 1 - Multiply:

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

Simplified the arguments

Calculated

Distributed

Example 1 - Divide

Example 1 - divide:

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