Simplify `(√3 + 3i)^31` Using De Moivre’s Theorem | Query Point Official
Simplify `\((\sqrt{3} + 3i)^{31}\)` Using De Moivre’s Theorem
Introduction to De Moivre’s Theorem
De Moivre’s theorem states that for any complex number written in polar form `\(z = r(\cos θ + i\sin θ)\)`,
`z^n = r^n [\cos(nθ) + i\sin(nθ)]`
This theorem simplifies raising complex numbers to powers by working via the modulus and argument instead of repeated multiplication.
Step 1: Convert to Polar Form
Given:
`z = (√3 + 3i)`
Find the modulus:
`r = √[(√3)^2 + (3)^2] = √(3 + 9) = √12`
Find the argument (θ):
`θ = tan^{−1}(3/√3) = tan^{−1}(√3) = π/3`
Step 2: Apply De Moivre’s Theorem
Raise z to the 31st power using De Moivre’s theorem:
`z^31 = (√12)^{31} [\cos(31·π/3) + i\sin(31·π/3)]`
Now simplify the angle:
`31·π/3 = (30 + 1)·π/3 = 10π + π/3`
Since cosine and sine have period 2π:
`cos(10π + π/3) = cos(π/3)` and `sin(10π + π/3) = sin(π/3)`
Step 3: Evaluate Final Result
Using known values:
- `cos(π/3) = 1/2`
- `sin(π/3) = √3/2`
So the simplified form is:
`(√3 + 3i)^{31} = (√12)^{31} [1/2 + i(√3/2)]`
This is the final simplified expression in polar form, using De Moivre’s theorem.
Explanation
De Moivre’s theorem reduces complex exponentiation to simpler operations on modulus and angle. Instead of multiplying the complex number by itself 31 times, we raise the modulus to the 31st power and multiply the argument by 31. This method is especially useful for large powers of complex numbers.
FAQ
Q1: What is De Moivre’s theorem used for?
It is used to simplify powers of complex numbers when written in polar form: `r(\cos θ + i\sin θ)^n = r^n(\cos nθ + i\sin nθ)`.
Q2: Why do we convert to polar form?
Polar form makes multiplication and exponentiation easier by separating magnitude and angle, avoiding repeated expansion.
Q3: How is the argument computed?
The argument θ of a complex number `a + bi` is found via `θ = tan^{-1}(b/a)` with quadrant considerations included.
Related Problems
See Mathematics Notes & MCQs for similar complex number problems and applications of De Moivre’s theorem.
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