De Moivre's Theorem

De Moirve's Theorem - How it Works - Video



These are the formulas that we will use to transform a complex number into standard form from trigonometric from or vice versa.



When we multiply complex numbers in trigonometric form, we multiply the modulus of z, r, and add the argument, θ.

The first line we have our original complex number.

The second line we have multiplied two r's and add another θ to both the cosine and sine so we have two for both.

The third line we have multiplied three r's and add another θ to both the cosine and sine so we have three for both.

The fourth line we have our formula zn = rn * (cos(nθ) + i * sin(nθ), where n is the number of times we multiply the same complex number.


Here the worked simplified.


Here is one reason why when we multiply two complex numbers in trigonometric form we multiply the modulus, r, and add the argument, θ.

Line 1 - we have one complex number.

Line 2 - we square both sides.

Line 3 - we distributed the terms.

Line 4 - we rearranged the terms.

Line 5 - we squared i.

Line 6 - we factored out an i.

Line 7 - we squared and combine like terms.

Line 8 - we use the double angle identities.

Example 1

Example 1:

Here we have our number in trigonometric 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.

42 + 42 = c2

16 + 16 = c2

32 = c2

sqrt(32) = c

sqrt(16) * sqrt(2) = c

c = 4 * sqrt(2)

Square the numbers


Square root both sides

Separate into two radicals

Square root one number

Now we know c, the hypotenuse, or our r, 4 * sqrt(2). Now, we must solve for our argument or theta, by using tanθ = b/a. Remember b is the opposite side length while a is the adjacent side length.

tanθ = b/a

tanθ = 4/4

tanθ = 1

θ = tan-1(1)

θ = π/4

Substitute 4 for a and 4 for b


Inverse tangent on both sides


Now, we have our r and theta. We can substitute into our De Moivre's Theorem.

z = [4 * sqrt(2)]5 * [cos(5 * π/4) + i * sin(5 * π/4)]

z = 4096 * sqrt(2) * [cos(5π/4) + i * sin(5π/4)]

z = 4096 * sqrt(2) * [(-sqrt(2)/2) + i * (-sqrt(2)/2)]

z = 4096 * sqrt(2) * (-sqrt(2)/2) + 4096 * sqrt(2) + i * (-sqrt(2)/2)]

z = -4096 - 4096i

Square the numbers

Simplify cosine and sine


Multiple terms

Now, we have complex number in standard form, a + bi, which is -4096 - 4096i.

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