MATH 1201 (Polar Coordinates - De Moivre\'s Theorem) Discussion post Unit 8 PDF

Preview

Summary

MATH 1201 (Polar Coordinates - De Moivre's Theorem) Discussion post Unit 8. Explanation of De Moivre's Theorem with sample questions and answers....

Description

De Moivre’s Theorem is a theorem that plays a great role in obtaining the formula for computing powers of complex roots. It states that, for a positive integer n, zn is found by raising the modulus to the nth power and multiplying the argument by n (Abramson, 2017, p. 821). If z = r (cos θ + isin θ) is a complex number, then zn = rn [cos(nθ) + isin(nθ)] zn = rn cis(nθ) where n is a positive integer. The scope of De Moivre's Theorem is to find the roots and powers of any complex numbers. This helps to find Trigonometric formulas easily enabling us to determine sine and cosine of multiple angles. It applies to complex numbers written in polar form such that any complex number z = a + ib can be represented in the form of z = r (cos θ + isin θ).

De Moivre's Theorem to Find Power of Complex Numbers If z is a complex number in polar form written as Z = r (cos(θ) + isin(θ)) then zn = rn(cos(nθ) + isin(nθ)) where n is an integer.

Example 1. i23 Write i in polar form i = cos(π / 2) + isin(π /2)

Use De Moivre's theorem to find i23

i23 = (cos(π / 2) + isin(π / 2))23 = (cos(23π / 2) + isin(23π / 2)) Simplify = cos(3π / 2 + 5(2π)) + isin(3π / 2 + 5(2π)) = cos(3π / 2) + isin(3π / 2) = −i

Example 2. (1−i)12 Write (1−i) in polar form (1−i) = 2–√(cos(7π / 4) + isin(7π / 4))

Use De Moivre's theorem to find (1−i)12

(1−i)12 = (2–√(cos(7π / 4) + isin(7π / 4)))12 = (2–√)12(cos(12 × 7π / 4) + isin(12 × 7π / 4))

Simplify = 64(cos(21π) + isin(21π)) = 64(−1 + 0) = −64

De Moivre's Theorem to Find Roots of Complex Numbers

De Moivre's theorem can also be used to find the nth roots of a complex number as follows If z is a complex number of the form Z = r(cos(θ) + isin(θ)) then the nth roots are given by zk = r1/n(cos(θ + 2kπn) + isin(θ + 2kπn)) where k = 0, 1, ... , (n - 1).

Example 1. All the third roots of 1 1 in polar form 1=1(cos(0) + isin(0))

Use the De Moivre's theorem to find all third roots

zk = 11/3(cos(0 + 2kπ / 3) + isin(0 + 2kπ / 3)) , k = 0, 1,2. The three roots are Set k = 0, 1 and 2 in the formula above to obtain the roots: z0 = 11/3(cos(0) + isin(0)) = 1 z1 = 11/3(cos(2π / 3) + isin(2π / 3)) = −12+i3√2 z2 = 11/3(cos(4π / 3) + isin(4π / 3) = −12−i3√2)

The three third roots are shown on the complex plane below. They are located on the same circle because their moduli are equal and they are equally spaced because the difference between their arguments are equal.

Example 2.

All the third roots of i i in polar form i = 1(cos(π / 2) + isin(π / 2))

Use De Moivre's theorem to find all third roots

zk=11/3(cos(π / 2 + 2kπ / 3) + isin(π/ 2 + 2kπ / 3)) , k = 0, 1,2. Set k = 0, 1 and 2 in the formula above to obtain the roots: z0 = 11/3(cos( π/ 6) + isin(π / 6)) = 3√2 + i12 z1 = 11/3(cos(π / 2 + 2π3) + isin(π / 2 + 2π3)) = (cos(5π / 6) + isin(5π / 6)) = −3√2 + i12 z2 = 11/3(cos(π / 2 + 4π3) + isin(π / 2 + 4π3)) = cos(3π / 2) + isin(3π / 2) = −i

References

Abramson, J. (2017). Algebra and trigonometry. OpenStax,TX: Rice University. Retrieved from https://openstax.org.details/algebra-and-trigonometry

Free mathematics tutorials, (n.d.) De Moivre's Theorem Power and Root https://www.analyzemath.com/complex/De-Moivre%27s-theorem-power-and-roots.html...